Đề thi thử tốt nghiệp THPT QG môn Toán năm 2020 - Bộ đề 13 Đề thi thử tốt nghiệp THPT Quốc gia môn Toán năm 2020, miễn phí và có đáp án đầy đủ. Nội dung bao gồm các dạng bài như giải tích, logarit, tích phân và bài toán thực tế.
Từ khoá: Toán học giải tích logarit tích phân bài toán thực tế năm 2020 đề thi thử tốt nghiệp đề thi có đáp án
Bộ sưu tập: 📘 Tuyển Tập Bộ 500 Đề Thi Ôn Luyện Môn Toán THPT Quốc Gia Các Tỉnh Từ Năm 2018-2025 - Có Đáp Án Chi Tiết 📘 Tuyển Tập Đề Thi Tham Khảo Các Môn THPT Quốc Gia 2025 🎯
Câu 1: Tính tích phân % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qCaeaada % qadaqaaiaaikdacaWGHbGaamiEaiabgUcaRiaadkgaaiaawIcacaGL % PaaacaqGKbGaamiEaaWcbaGaaGymaaqaaiaaikdaa0Gaey4kIipaaa % a!41A8! \int\limits_1^2 {\left( {2ax + b} \right){\rm{d}}x}
Câu 2: Tính đạo hàm f'(x) của hàm số % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9iGacYgacaGGVbGaai4z % amaaBaaaleaacaaIYaaabeaakmaabmaabaGaaG4maiaadIhacqGHsi % slcaaIXaaacaGLOaGaayzkaaaaaa!4318! f\left( x \right) = {\log _2}\left( {3x - 1} \right)\)</span> với <span class="math-tex">\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg6 % da+maalaaabaGaaGymaaqaaiaaiodaaaGaaiOlaaaa!3A33! x > \frac{1}{3}.
A. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafa % WaaeWaaeaacaWG4baacaGLOaGaayzkaaGaeyypa0ZaaSaaaeaacaaI % ZaGaciiBaiaac6gacaaIYaaabaWaaeWaaeaacaaIZaGaamiEaiabgk % HiTiaaigdaaiaawIcacaGLPaaaaaaaaa!42CF! f'\left( x \right) = \frac{{3\ln 2}}{{\left( {3x - 1} \right)}} B. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafa % WaaeWaaeaacaWG4baacaGLOaGaayzkaaGaeyypa0ZaaSaaaeaacaaI % XaaabaWaaeWaaeaacaaIZaGaamiEaiabgkHiTiaaigdaaiaawIcaca % GLPaaaciGGSbGaaiOBaiaaikdaaaaaaa!42CD! f'\left( x \right) = \frac{1}{{\left( {3x - 1} \right)\ln 2}} C. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafa % WaaeWaaeaacaWG4baacaGLOaGaayzkaaGaeyypa0ZaaSaaaeaacaaI % ZaaabaWaaeWaaeaacaaIZaGaamiEaiabgkHiTiaaigdaaiaawIcaca % GLPaaaaaaaaa!402F! f'\left( x \right) = \frac{3}{{\left( {3x - 1} \right)}} D. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafa % WaaeWaaeaacaWG4baacaGLOaGaayzkaaGaeyypa0ZaaSaaaeaacaaI % ZaaabaWaaeWaaeaacaaIZaGaamiEaiabgkHiTiaaigdaaiaawIcaca % GLPaaaciGGSbGaaiOBaiaaikdaaaaaaa!42CF! f'\left( x \right) = \frac{3}{{\left( {3x - 1} \right)\ln 2}}
Câu 3: Người ta muốn mạ vàng cho một cái hộp có đáy hình vuông không nắp có thể tích là 4 lít. Tìm kích thước của hộp đó để lượng vàng dùng mạ là ít nhất. Giả sử độ dày của lớp mạ tại mọi nơi trên mặt ngoài hộp là như nhau.
A. Cạnh đáy bằng 1, chiều cao bằng 2
B. Cạnh đáy bằng 4, chiều cao bằng 3.
C. Cạnh đáy bằng 2, chiều cao bằng 1.
D. Cạnh đáy bằng 3, chiều cao bằng 4.
Câu 4: Hàm số y = f(x)\) liên tục và có bảng biến thiên trong đoạn \([-1;3]\) cho trong hình bên. Gọi M là giá trị lớn nhất của hàm số \(f(x)\) trên đoạn \([-1;3]
Câu 5: Trong không gian với hệ tọa độ Oxyz cho đường thẳng % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8srps0l % bbf9q8WrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0-yr0R % Yxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa % caGaaeqabaqaaeaadaaakeaacaWGKbGaaiOoamaalaaabaGaamiEai % abgUcaRiaaiodaaeaacaaIYaaaaiabg2da9maalaaabaGaamyEaiab % gkHiTiaaigdaaeaacaaIXaaaaiabg2da9maalaaabaGaamOEaiabgk % HiTiaaigdaaeaacqGHsislcaaIZaaaaaaa!40A4! d:\frac{{x + 3}}{2} = \frac{{y - 1}}{1} = \frac{{z - 1}}{{ - 3}}
A. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WG1baacaGLxdcacqGH9aqpdaqadaqaaiaaikdacaGG7aGaaGymaiaa % cUdacqGHsislcaaIZaaacaGLOaGaayzkaaaaaa!3FD0! \overrightarrow u = \left( {2;1; - 3} \right) B. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WG1baacaGLxdcacqGH9aqpdaqadaqaaiaaikdacaGG7aGaaGimaiaa % cUdacaaIWaaacaGLOaGaayzkaaaaaa!3EDF! \overrightarrow u = \left( {2;0;0} \right) C. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WG1baacaGLxdcacqGH9aqpdaqadaqaaiaaicdacaGG7aGaaGymaiaa % cUdacaaIZaaacaGLOaGaayzkaaaaaa!3EE1! \overrightarrow u = \left( {0;1;3} \right) D. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WG1baacaGLxdcacqGH9aqpdaqadaqaaiaaicdacaGG7aGaaGymaiaa % cUdacqGHsislcaaIZaaacaGLOaGaayzkaaaaaa!3FCE! \overrightarrow u = \left( {0;1; - 3} \right)
Câu 6: Cho hàm số % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWG5bGaeyypa0ZaaSaaaeaacaWG4bGaey4kaSIaaGymaaqaaiaa % dIhacqGHsislcaaIYaaaaiaaywW7caGGOaGaam4qaiaacMcaaaa!4116! y = \frac{{x + 1}}{{x - 2}}\quad (C)
A. % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaGcaaqaaiaaiodaaSqabaaaaa!36EB! \sqrt 3 B. % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaGcaaqaaiaaiAdaaSqabaaaaa!36EE! \sqrt 6 C. % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaWcaaqaamaakaaabaGaaGOmaaWcbeaaaOqaaiaaikdaaaaaaa!37C0! \frac{{\sqrt 2 }}{2} D. % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaGcaaqaaiaaiwdaaSqabaaaaa!36ED! \sqrt 5
Câu 7: Trong không gian với hệ tọa độ Oxyx , cho đường thẳng % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaacQ % dadaWcaaqaaiaadIhacqGHsislcaaIXaaabaGaaGymaaaacqGH9aqp % daWcaaqaaiaadMhacqGHsislcaaIYaaabaGaaGymaaaacqGH9aqpda % WcaaqaaiaadQhacqGHsislcaaIXaaabaGaaGOmaaaaaaa!43FB! d:\frac{{x - 1}}{1} = \frac{{y - 2}}{1} = \frac{{z - 1}}{2}\)</span>,A(2;1;4) . Gọi H(a;b;c) là điểm thuộc d sao cho AH có độ dài nhỏ nhất. Tính <span class="math-tex">\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivaiabg2 % da9iaadggadaahaaWcbeqaaiaaiodaaaGccqGHRaWkcaWGIbWaaWba % aSqabeaacaaIZaaaaOGaey4kaSIaam4yamaaCaaaleqabaGaaG4maa % aaaaa!3F1D! T = {a^3} + {b^3} + {c^3}
B. % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivaiabg2 % da9maakaaabaGaaGynaaWcbeaaaaa!38AC! T = \sqrt 5
Câu 8: Gọi z_0\) là nghiệm phức có phần ảo âm của phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfeaY-biLkVcLq-JHqpepeea0-as0Fb9pgeaYRXxe9vr0-vr % 0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaaikdacaWG6b % WaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaGOnaiaadQhacqGHRaWk % caaI1aGaeyypa0JaaGimaaaa!3EA3! 2{z^2} - 6z + 5 = 0\) . Số phức \(iz_0
A. % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfeaY-biLkVcLq-JHqpepeea0-as0Fb9pgeaYRXxe9vr0-vr % 0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaaG % ymaaqaaiaaikdaaaGaeyOeI0YaaSaaaeaacaaIZaaabaGaaGOmaaaa % caWGPbaaaa!3AD3! \frac{1}{2} - \frac{3}{2}i B. % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfeaY-biLkVcLq-JHqpepeea0-as0Fb9pgeaYRXxe9vr0-vr % 0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabgkHiTmaala % aabaGaaGymaaqaaiaaikdaaaGaey4kaSYaaSaaaeaacaaIZaaabaGa % aGOmaaaacaWGPbaaaa!3BB5! - \frac{1}{2} + \frac{3}{2}i C. % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfeaY-biLkVcLq-JHqpepeea0-as0Fb9pgeaYRXxe9vr0-vr % 0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaaG % ymaaqaaiaaikdaaaGaey4kaSYaaSaaaeaacaaIZaaabaGaaGOmaaaa % caWGPbaaaa!3AC8! \frac{1}{2} + \frac{3}{2}i D. % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfeaY-biLkVcLq-JHqpepeea0-as0Fb9pgeaYRXxe9vr0-vr % 0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabgkHiTmaala % aabaGaaGymaaqaaiaaikdaaaGaeyOeI0YaaSaaaeaacaaIZaaabaGa % aGOmaaaacaWGPbaaaa!3BC0! - \frac{1}{2} - \frac{3}{2}i
Câu 9: Trong không gian với hệ trục tọa độ Oxyz , gọi % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaacq % aHXoqyaiaawIcacaGLPaaaaaa!391C! \left( \alpha \right)\)</span> là mặt phẳng chứa đường thẳng <span class="math-tex">\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdqKaai % OoaiaaykW7daWcaaqaaiaadIhacqGHsislcaaIYaaabaGaaGymaaaa % cqGH9aqpdaWcaaqaaiaadMhacqGHsislcaaIXaaabaGaaGymaaaacq % GH9aqpdaWcaaqaaiaadQhaaeaacqGHsislcaaIYaaaaaaa!4549! \Delta :\,\frac{{x - 2}}{1} = \frac{{y - 1}}{1} = \frac{z}{{ - 2}}\)</span> và vuông góc với mặt phẳng <span class="math-tex">\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaacq % aHYoGyaiaawIcacaGLPaaacaGG6aGaaGPaVlaadIhacqGHRaWkcaWG % 5bGaey4kaSIaaGOmaiaadQhacqGHRaWkcaaIXaGaeyypa0JaaGimaa % aa!443E! \left( \beta \right):\,x + y + 2z + 1 = 0\)</span>. Khi đó giao tuyến của hai mặt phẳng <span class="math-tex">\((\alpha) ; % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaacq % aHYoGyaiaawIcacaGLPaaaaaa!391E! \left( \beta \right)
A. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGPaVpaala % aabaGaamiEaaqaaiaaigdaaaGaeyypa0ZaaSaaaeaacaWG5bGaey4k % aSIaaGymaaqaaiaaigdaaaGaeyypa0ZaaSaaaeaacaWG6baabaGaey % OeI0IaaGymaaaaaaa!4170! \,\frac{x}{1} = \frac{{y + 1}}{1} = \frac{z}{{ - 1}} B. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WG4baabaGaaGymaaaacqGH9aqpdaWcaaqaaiaadMhacqGHRaWkcaaI % XaaabaGaaGymaaaacqGH9aqpdaWcaaqaaiaadQhacqGHsislcaaIXa % aabaGaaGymaaaaaaa!40A0! \frac{x}{1} = \frac{{y + 1}}{1} = \frac{{z - 1}}{1} C. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WG4bGaeyOeI0IaaGOmaaqaaiaaigdaaaGaeyypa0ZaaSaaaeaacaWG % 5bGaey4kaSIaaGymaaqaaiabgkHiTiaaiwdaaaGaeyypa0ZaaSaaae % aacaWG6baabaGaaGOmaaaaaaa!4193! \frac{{x - 2}}{1} = \frac{{y + 1}}{{ - 5}} = \frac{z}{2} D. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WG4bGaey4kaSIaaGOmaaqaaiaaigdaaaGaeyypa0ZaaSaaaeaacaWG % 5bGaeyOeI0IaaGymaaqaaiabgkHiTiaaiwdaaaGaeyypa0ZaaSaaae % aacaWG6baabaGaaGOmaaaaaaa!4193! \frac{{x + 2}}{1} = \frac{{y - 1}}{{ - 5}} = \frac{z}{2}
Câu 10: Cho hàm số % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9Fve9 % Ff0dmeGabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWG5bGaeyypa0 % ZaaSaaaeaacaWG4bGaeyOeI0IaaGymaaqaaiaaikdacqGHsislcaWG % 4baaaaaa!3CE1! y = \frac{{x - 1}}{{2 - x}}
A. % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9Fve9 % Ff0dmeGabaqaciGacaGaaeqabaWaaeaaeaaakeaacqGHsisldaWcaa % qaaiaaiodaaeaacaaIYaaaaaaa!37F8! - \frac{3}{2} C. % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9Fve9 % Ff0dmeGabaqaciGacaGaaeqabaWaaeaaeaaakeaacqGHsisldaWcaa % qaaiaaiwdaaeaacaaIYaaaaaaa!37FA! - \frac{5}{2}
Câu 11: Tìm nguyên hàm của hàm số f(x) = 2x + 1
A. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qaaeaada % qadaqaaiaaikdacaWG4bGaey4kaSIaaGymaaGaayjkaiaawMcaaaWc % beqab0Gaey4kIipakiaabsgacaWG4bGaeyypa0ZaaSaaaeaacaWG4b % WaaWbaaSqabeaacaaIYaaaaaGcbaGaaGOmaaaacqGHRaWkcaWG4bGa % ey4kaSIaam4qaaaa!4607! \int {\left( {2x + 1} \right)} {\rm{d}}x = \frac{{{x^2}}}{2} + x + C B. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qaaeaada % qadaqaaiaaikdacaWG4bGaey4kaSIaaGymaaGaayjkaiaawMcaaaWc % beqab0Gaey4kIipakiaabsgacaWG4bGaeyypa0JaamiEamaaCaaale % qabaGaaGOmaaaakiabgUcaRiaadIhacqGHRaWkcaWGdbaaaa!453B! \int {\left( {2x + 1} \right)} {\rm{d}}x = {x^2} + x + C C. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qaaeaada % qadaqaaiaaikdacaWG4bGaey4kaSIaaGymaaGaayjkaiaawMcaaaWc % beqab0Gaey4kIipakiaabsgacaWG4bGaeyypa0JaaGOmaiaadIhada % ahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaIXaGaey4kaSIaam4qaaaa % !45B5! \int {\left( {2x + 1} \right)} {\rm{d}}x = 2{x^2} + 1 + C D. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qaaeaada % qadaqaaiaaikdacaWG4bGaey4kaSIaaGymaaGaayjkaiaawMcaaaWc % beqab0Gaey4kIipakiaabsgacaWG4bGaeyypa0JaamiEamaaCaaale % qabaGaaGOmaaaakiabgUcaRiaadoeaaaa!435C! \int {\left( {2x + 1} \right)} {\rm{d}}x = {x^2} + C
Câu 12: Cho hàm số y = f(x). Hàm số y = f'(x) có đồ thị như hình vẽ
Hàm số % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaadAgadaqadaqaaiaadIhadaahaaWcbeqaaiaaikdaaaaakiaa % wIcacaGLPaaaaaa!3C5C! y = f\left( {{x^2}} \right)
Câu 13: Có bao nhiêu số hạng trong khai triển nhị thức % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % aIYaGaamiEaiabgkHiTiaaiodaaiaawIcacaGLPaaadaahaaWcbeqa % aiaaikdacaaIWaGaaGymaiaaiIdaaaaaaa!3E00! {\left( {2x - 3} \right)^{2018}}
Câu 14: Số mặt cầu chứa một đường tròn cho trước là
Câu 15: Cho hình chóp S.ABCD có đáy ABCD là hình vuông tâm O cạnh a, SO vuông góc với mặt phẳng (ABCD) và SO = a. Khoảng cách giữa SC và AB bằng
A. % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIYaGaamyyamaakaaabaGaaGynaaWcbeaaaOqaaiaaiwdaaaaaaa!3948! \frac{{2a\sqrt 5 }}{5} B. % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WGHbWaaOaaaeaacaaI1aaaleqaaaGcbaGaaGynaaaaaaa!388C! \frac{{a\sqrt 5 }}{5} C. % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIYaGaamyyamaakaaabaGaaG4maaWcbeaaaOqaaiaaigdacaaI1aaa % aaaa!3A01! \frac{{2a\sqrt 3 }}{{15}} D. % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIYaGaamyyamaakaaabaGaaG4maaWcbeaaaOqaaiaaigdacaaI1aaa % aaaa!3A01! \frac{{a\sqrt 3 }}{{15}}
Câu 16: Diện tích hình phẳng giới hạn bởi đồ thị hàm số % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Lq-Jirpepeea0-as0Fb9pgea0lXxe9vr0-vr % 0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadMhacqGH9a % qpdaWcaaqaaiaadIhacqGHRaWkcaaIXaaabaGaamiEaiabgkHiTiaa % ikdaaaaaaa!3DBB! y = \frac{{x + 1}}{{x - 2}}
A. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Lq-Jirpepeea0-as0Fb9pgea0lXxe9vr0-vr % 0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaaiodaciGGSb % GaaiOBamaalaaabaGaaGynaaqaaiaaikdaaaGaeyOeI0IaaGymaaaa % !3C3B! 3\ln \frac{5}{2} - 1 B. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Lq-Jirpepeea0-as0Fb9pgea0lXxe9vr0-vr % 0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaaikdaciGGSb % GaaiOBamaalaaabaGaaG4maaqaaiaaikdaaaGaeyOeI0IaaGymaaaa % !3C38! 2\ln \frac{3}{2} - 1 C. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Lq-Jirpepeea0-as0Fb9pgea0lXxe9vr0-vr % 0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaaiwdaciGGSb % GaaiOBamaalaaabaGaaG4maaqaaiaaikdaaaGaeyOeI0IaaGymaaaa % !3C3B! 5\ln \frac{3}{2} - 1 D. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Lq-Jirpepeea0-as0Fb9pgea0lXxe9vr0-vr % 0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaaiodaciGGSb % GaaiOBamaalaaabaGaaG4maaqaaiaaikdaaaGaeyOeI0IaaGymaaaa % !3C39! 3\ln \frac{3}{2} - 1
Câu 17: Một hình nón có chiều cao bằng a 3 a\sqrt 3 a 3
A. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uamaaBa % aaleaacaWG4bGaamyCaaqabaGccqGH9aqpcqaHapaCcaWGHbWaaWba % aSqabeaacaaIYaaaaaaa!3D87! {S_{xq}} = \pi {a^2}\)</span><span class="math-tex">\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uamaaBa % aaleaacaWG4bGaamyCaaqabaGccqGH9aqpcqaHapaCcaWGHbWaaWba % aSqabeaacaaIYaaaaaaa!3D87! {S_{xq}} = \pi {a^2} B. S x q = 2 π a 2 {S_{xq}} = 2\pi {a^2} S x q = 2 π a 2 C. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uamaaBa % aaleaacaWG4bGaamyCaaqabaGccqGH9aqpdaGcaaqaaiaaiodaaSqa % baGccqaHapaCcaWGHbWaaWbaaSqabeaacaaIYaaaaaaa!3E69! {S_{xq}} = \sqrt 3 \pi {a^2} D. S x q = 2 a 2 {S_{xq}} = 2{a^2} S x q = 2 a 2
Câu 18: Cho hai số phức % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOEamaaBa % aaleaacaaIXaaabeaakiabg2da9iaaikdacqGHRaWkcaaIZaGaamyA % aaaa!3C33! {z_1} = 2 + 3i\)</span>,<span class="math-tex">\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOEamaaBa % aaleaacaaIYaaabeaakiabg2da9iabgkHiTiaaisdacqGHsislcaaI % 1aGaamyAaaaa!3D30! {z_2} = - 4 - 5i\)</span> . Số phức <span class="math-tex">\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOEaiabg2 % da9iaadQhadaWgaaWcbaGaaGymaaqabaGccqGHRaWkcaWG6bWaaSba % aSqaaiaaikdaaeqaaaaa!3CB2! z = {z_1} + {z_2}
Câu 19: Cho hình tứ diện OABC có đáy OBC là tam giác vuông tại O,OB =a ,OC= a\sqrt3\) . Cạnh OA vuông góc với mặt phẳng (OBC), \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4taiaadg % eacqGH9aqpcaWGHbWaaOaaaeaacaaIZaaaleqaaaaa!3A52! OA = a\sqrt 3
A. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiAaiabg2 % da9maalaaabaGaamyyamaakaaabaGaaGymaiaaiwdaaSqabaaakeaa % caaI1aaaaaaa!3B3B! h = \frac{{a\sqrt {15} }}{5} B. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiAaiabg2 % da9maalaaabaGaamyyamaakaaabaGaaG4maaWcbeaaaOqaaiaaikda % aaaaaa!3A7B! h = \frac{{a\sqrt 3 }}{2} C. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiAaiabg2 % da9maalaaabaGaamyyamaakaaabaGaaG4maaWcbeaaaOqaaiaaigda % caaI1aaaaaaa!3B39! h = \frac{{a\sqrt 3 }}{{15}} D. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiAaiabg2 % da9maalaaabaGaamyyamaakaaabaGaaGynaaWcbeaaaOqaaiaaiwda % aaaaaa!3A80! h = \frac{{a\sqrt 5 }}{5}
Câu 20: Với điều kiện % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaeaafa % qabeGabaaabaGaamyyaiaadogacaGGOaGaamOyamaaCaaaleqabaGa % aGOmaaaakiabgkHiTiaaisdacaWGHbGaam4yaiaacMcacqGH+aGpca % aIWaaabaGaamyyaiaadkgacqGH8aapcaaIWaaaaaGaay5Eaaaaaa!44E2! \left\{ {\begin{array}{*{20}{c}} {ac({b^2} - 4ac) > 0}\\ {ab < 0} \end{array}} \right.\)</span> thì đồ thị hàm số <span class="math-tex">\(y = ax^4+bx^2+c
Câu 21: Tính diện tích miền hình phẳng giới hạn bởi các đường % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaadIhadaahaaWcbeqaaiaaikdaaaGccqGHsislcaaIYaGaamiE % aaaa!3C8E! y = {x^2} - 2x
A. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiabg2 % da9maalaaabaGaaGOmaiaaicdacaaIWaGaaGimaaqaaiaaiodaaaaa % aa!3B89! S = \frac{{2000}}{3} C. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiabg2 % da9maalaaabaGaaGOmaiaaicdacaaIWaGaaGioaaqaaiaaiodaaaaa % aa!3B91! S = \frac{{2008}}{3}
Câu 22: Gọi M là điểm biểu diễn của số phức z trong mặt phẳng tọa độ, N là điểm đối xứng của M qua Oy (M ,N không thuộc các trục tọa độ). Số phức w có điểm biểu diễn lên mặt phẳng tọa độ là N. Mệnh đề nào sau đây đúng ?
B. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4Daiabg2 % da9iabgkHiTiqadQhagaqeaaaa!39FA! w = - \bar z C. % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4Daiabg2 % da9iqadQhagaqeaaaa!390C! w = \bar z D. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaaca % WG3baacaGLhWUaayjcSdGaeyOpa4ZaaqWaaeaacaWG6baacaGLhWUa % ayjcSdaaaa!3F3B! \left| w \right| > \left| z \right|
Câu 23: Số giá trị nguyên của m < 0 để hàm số % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iGacYgacaGGUbWaaeWaaeaacaWG4bWaaWbaaSqabeaacaaIYaaa % aOGaey4kaSIaamyBaiaadIhacqGHRaWkcaaIXaaacaGLOaGaayzkaa % aaaa!41C3! y = \ln \left( {{x^2} + mx + 1} \right)\)</span> đồng biến trên <span class="math-tex">\((0;+\infty)
Câu 24: Cho hàm số % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaadIhadaahaaWcbeqaaiaaiodaaaGccqGHsislcaaIZaGaamiE % amaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaiodacaWGTbGaamiEai % abgUcaRiaad2gacqGHsislcaaIXaaaaa!448C! y = {x^3} - 3{x^2} + 3mx + m - 1
Câu 25: Trong không gian Oxyz, cho hình thoi ABCD với A(-1;2;1) ; B (2;3;2). Tâm I của hình thoi thuộc đường thẳng % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaacQ % dadaWcaaqaaiaadIhacqGHRaWkcaaIXaaabaGaeyOeI0IaaGymaaaa % cqGH9aqpdaWcaaqaaiaadMhaaeaacqGHsislcaaIXaaaaiabg2da9m % aalaaabaGaamOEaiabgkHiTiaaikdaaeaacaaIXaaaaaaa!4421! d:\frac{{x + 1}}{{ - 1}} = \frac{y}{{ - 1}} = \frac{{z - 2}}{1}
Câu 26: Cho đồ thị hàm số y = f(x) có đồ thị như hình vẽ. Hàm số y = f(x) đồng biến trên khoảng nào dưới đây?
B. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGabiqaaiaabeqaamaabaabaaGcbaWaaeWaaeaacq % GHsislcaaMc8UaeyOhIuQaai4oaiaaykW7caaMc8UaaGimaaGaayjk % aiaawMcaaaaa!3FF3! \left( { - \,\infty ;\,\,0} \right) D. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGabiqaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % aIYaGaai4oaiaaykW7caaMc8Uaey4kaSIaeyOhIukacaGLOaGaayzk % aaaaaa!3E5F! \left( {2;\,\, + \infty } \right)
Câu 27: Cho f,g là hai hàm liên tục trên [1;3] thỏa điều kiện % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Lq-Jirpepeea0-as0Fb9pgea0lXxe9vr0-vr % 0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaapehabaWaam % WaaeaacaWGMbWaaeWaaeaacaWG4baacaGLOaGaayzkaaGaey4kaSIa % aG4maiaadEgadaqadaqaaiaadIhaaiaawIcacaGLPaaaaiaawUfaca % GLDbaacaqGKbGaamiEaaWcbaGaaGymaaqaaiaaiodaa0Gaey4kIipa % kiabg2da9iaaigdacaaIWaaaaa!4925! \int\limits_1^3 {\left[ {f\left( x \right) + 3g\left( x \right)} \right]{\rm{d}}x} = 10\)</span> đồng thời <span class="math-tex">\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Lq-Jirpepeea0-as0Fb9pgea0lXxe9vr0-vr % 0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaapehabaWaam % WaaeaacaaIYaGaamOzamaabmaabaGaamiEaaGaayjkaiaawMcaaiab % gkHiTiaadEgadaqadaqaaiaadIhaaiaawIcacaGLPaaaaiaawUfaca % GLDbaacaqGKbGaamiEaaWcbaGaaGymaaqaaiaaiodaa0Gaey4kIipa % kiabg2da9iaaiAdaaaa!4879! \int\limits_1^3 {\left[ {2f\left( x \right) - g\left( x \right)} \right]{\rm{d}}x} = 6\)</span>. Tính <span class="math-tex">\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Lq-Jirpepeea0-as0Fb9pgea0lXxe9vr0-vr % 0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaapehabaWaam % WaaeaacaWGMbWaaeWaaeaacaWG4baacaGLOaGaayzkaaGaey4kaSIa % am4zamaabmaabaGaamiEaaGaayjkaiaawMcaaaGaay5waiaaw2faai % aabsgacaWG4baaleaacaaIXaaabaGaaG4maaqdcqGHRiI8aaaa!45E3! \int\limits_1^3 {\left[ {f\left( x \right) + g\left( x \right)} \right]{\rm{d}}x}
Câu 28: Nghiệm của phương trình % MathType!MTEF!2!1!+- % feaahqart1ev3aqaM5cvLHfij5gC1rhimfMBNvxyNvga7TNm951EYG % xlX0xFTWLzYf2y7ftF7HtF9adatCvAUfeBSjuyZL2yd9gzLbvyNv2C % aerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLD % harqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8qqaqFr % 0xc9pk0xbba9q8WqFfeaY-biLkVcLq-JHqpepeea0-as0Fb9pgeaYR % Xxe9vr0-vr0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaaba % aaaaaaaapeGaaGOma8aadaahaaWcbeqaa8qacaaIYaGaamiEaiabgk % HiTiaaigdaaaGccqGHsisldaWcaaWdaeaapeGaaGymaaWdaeaapeGa % aGioaaaacqGH9aqpcaaIWaaaaa!4F78! {2^{2x - 1}} - \frac{1}{8} = 0
Câu 29: Hàm số % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaadIhadaahaaWcbeqaaiaaisdaaaGccqGHRaWkcaaIYaGaamiE % amaaCaaaleqabaGaaGOmaaaakiabgkHiTiaaiodaaaa!3F22! y = {x^4} + 2{x^2} - 3
Câu 30: Cho hàm số % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maalaaabaGaamiEaiabgUcaRiaaikdaaeaacaWG4bGaey4kaSIa % aGymaaaaaaa!3D3D! y = \frac{{x + 2}}{{x + 1}}
A. % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaaIZaWaaOaaaeaacaaIZaaaleqaaaaa!37A8! 3\sqrt 3 C. % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaaIZaWaaOaaaeaacaaIZaaaleqaaaaa!37A8! \sqrt 3
Câu 31: Cho hàm số y = f(x) có bảng biến thiên như sau:
Mệnh đề nào dưới đây đúng?
A. Hàm số nghịch biến trên khoảng (-1;1).
B. Hàm số đồng biến trên khoảng
% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGacaGaaiaabaqaamaabaabaaGcbaWaaeWaaeaacq % GHsislcqGHEisPcaGG7aGaaGjbVlaaigdaaiaawIcacaGLPaaaaaa!3CDF! (- \infty ;\;1) .
C. Hàm số nghịch biến trên khoảng (-1;3).
D. Hàm số đồng biến trên khoảng
( − 1 ; + ∞ ) (-1 ; +\infty) ( − 1 ; + ∞ ) .
Câu 32: Cho hình chóp S.ABCD có đáy ABCD là hình vuông cạnh a, tam giác SAB đều và nằm trong mặt phẳng vuông góc với đáy. Tính thể tích khối cầu ngoại tiếp khối chóp SABCD.
A. 7 21 54 π a 3 \frac{{7\sqrt {21} }}{{54}}\pi {a^3} 54 7 21 π a 3 B. 7 21 162 π a 3 \frac{{7\sqrt {21} }}{{162}}\pi {a^3} 162 7 21 π a 3 C. 7 21 216 π a 3 \frac{{7\sqrt {21} }}{{216}}\pi {a^3} 216 7 21 π a 3 D. 49 21 36 π a 3 \frac{{49\sqrt {21} }}{{36}}\pi {a^3} 36 49 21 π a 3
Câu 33: Phương trình % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGOmamaaCa % aaleqabaGaamiEamaaCaaameqabaGaaGOmaaaaliabgkHiTiaaioda % caWG4bGaey4kaSIaaGOmaaaakiabg2da9iaaisdaaaa!3EE2! {2^{{x^2} - 3x + 2}} = 4\)</span> có 2 nghiệm là <span class="math-tex">\(x_1;x_2\)</span> . Hãy tính giá trị của <span class="math-tex">\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivaiabg2 % da9iaadIhadaqhaaWcbaGaaGymaaqaaiaaiodaaaGccqGHRaWkcaWG % 4bWaa0baaSqaaiaaikdaaeaacaaIZaaaaaaa!3E04! T = x_1^3 + x_2^3
Câu 34: Bất phương trình % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac+ % gacaGGNbWaaSbaaSqaaiaaikdaaeqaaOWaaSaaaeaacaWG4bWaaWba % aSqabeaacaaIYaaaaOGaeyOeI0IaaGOnaiaadIhacqGHRaWkcaaI4a % aabaGaaGinaiaadIhacqGHsislcaaIXaaaaiabgwMiZkaaicdaaaa!45E6! {\log _2}\frac{{{x^2} - 6x + 8}}{{4x - 1}} \ge 0\)</span> có tập nghiệm là <span class="math-tex">\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivaiabg2 % da9maajadabaWaaSaaaeaacaaIXaaabaGaaGinaaaacaGG7aGaamyy % aaGaayjkaiaaw2faaiabgQIiipaajibabaGaamOyaiaacUdacqGHRa % WkcqGHEisPaiaawUfacaGLPaaaaaa!445E! T = \left( {\frac{1}{4};a} \right] \cup \left[ {b; + \infty } \right)
Câu 35: Tập hợp tất cả các giá trị của m để phương trình % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEamaaCa % aaleqabaGaaGOmaaaakiabgUcaRiaad2gacaWG4bGaeyOeI0IaamyB % aiabgUcaRiaaigdacqGH9aqpcaaIWaaaaa!3FF0! {x^2} + mx - m + 1 = 0
A. % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaKGeaeaaca % aIXaGaai4oaiabgUcaRiabg6HiLcGaay5waiaawMcaaaaa!3B93! \left[ {1; + \infty } \right) B. % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % aIXaGaai4oaiabgUcaRiabg6HiLcGaayjkaiaawMcaaaaa!3B49! \left( {1; + \infty } \right) D. % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaacq % GHsislcaaIYaGaey4kaSYaaOaaaeaacaaI4aaaleqaaOGaai4oaiab % gUcaRiabg6HiLcGaayjkaiaawMcaaaaa!3E00! \left( { - 2 + \sqrt 8 ; + \infty } \right)
Câu 36: Mặt phẳng đi qua ba điểm A( 0 ; 0 ;2), B( 1 ; 0 ; 0 ) và C( 0 ; 3 ; 0) có phương trình là:
A. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WG4baabaGaaGymaaaacqGHRaWkdaWcaaqaaiaadMhaaeaacaaIZaaa % aiabgUcaRmaalaaabaGaamOEaaqaaiaaikdaaaGaeyypa0JaaGymaa % aa!3ED7! \frac{x}{1} + \frac{y}{3} + \frac{z}{2} = 1 B. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WG4baabaGaaGymaaaacqGHRaWkdaWcaaqaaiaadMhaaeaacaaIZaaa % aiabgUcaRmaalaaabaGaamOEaaqaaiaaikdaaaGaeyypa0JaeyOeI0 % IaaGymaaaa!3FC4! \frac{x}{1} + \frac{y}{3} + \frac{z}{2} = - 1 C. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WG4baabaGaaGOmaaaacqGHRaWkdaWcaaqaaiaadMhaaeaacaaIXaaa % aiabgUcaRmaalaaabaGaamOEaaqaaiaaiodaaaGaeyypa0JaaGymaa % aa!3ED7! \frac{x}{2} + \frac{y}{1} + \frac{z}{3} = 1 D. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WG4baabaGaaGOmaaaacqGHRaWkdaWcaaqaaiaadMhaaeaacaaIXaaa % aiabgUcaRmaalaaabaGaamOEaaqaaiaaiodaaaGaeyypa0JaeyOeI0 % IaaGymaaaa!3FC4! \frac{x}{2} + \frac{y}{1} + \frac{z}{3} = - 1
Câu 37: Tìm tất cả các giá trị thực của tham số a ( a > 0) thỏa mãn % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqr1ngB % PrgifHhDYfgasaacH8srps0lbbf9q8WrFfeuY-Hhbba9q8qqaqFr0x % c9pk0xbba9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8fr % Fve9Fve9Ff0dmeaabiqaceGabiGaaeaabaWaaeaaeaaakeaadaqada % qaaiaaikdadaahaaWcbeqaaiaadggaaaGccqGHRaWkdaWcaaqaaiaa % igdaaeaacaaIYaWaaWbaaSqabeaacaWGHbaaaaaaaOGaayjkaiaawM % caamaaCaaaleqabaGaaGOmaiaaicdacaaIXaGaaG4naaaakiabgsMi % JoaabmaabaGaaGOmamaaCaaaleqabaGaaGOmaiaaicdacaaIXaGaaG % 4naaaakiabgUcaRmaalaaabaGaaGymaaqaaiaaikdadaahaaWcbeqa % aiaaikdacaaIWaGaaGymaiaaiEdaaaaaaaGccaGLOaGaayzkaaWaaW % baaSqabeaacaWGHbaaaaaa!4F2D! {\left( {{2^a} + \frac{1}{{{2^a}}}} \right)^{2017}} \le {\left( {{2^{2017}} + \frac{1}{{{2^{2017}}}}} \right)^a}
A. % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqr1ngB % PrgifHhDYfgasaacH8srps0lbbf9q8WrFfeuY-Hhbba9q8qqaqFr0x % c9pk0xbba9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8fr % Fve9Fve9Ff0dmeaabiqaceGabiGaaeaabaWaaeaaeaaakeaacaaIWa % GaeyipaWJaamyyaiabgsMiJkaaikdacaaIWaGaaGymaiaaiEdaaaa!3E9F! 0 < a \le 2017 B. % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqr1ngB % PrgifHhDYfgasaacH8srps0lbbf9q8WrFfeuY-Hhbba9q8qqaqFr0x % c9pk0xbba9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8fr % Fve9Fve9Ff0dmeaabiqaceGabiGaaeaabaWaaeaaeaaakeaacaaIXa % GaeyipaWJaamyyaiabgYda8iaaikdacaaIWaGaaGymaiaaiEdaaaa!3DEF! 1 < a < 2017 C. % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqr1ngB % PrgifHhDYfgasaacH8srps0lbbf9q8WrFfeuY-Hhbba9q8qqaqFr0x % c9pk0xbba9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8fr % Fve9Fve9Ff0dmeaabiqaceGabiGaaeaabaWaaeaaeaaakeaacaWGHb % GaeyyzImRaaGOmaiaaicdacaaIXaGaaG4naaaa!3CF2! a \ge 2017 D. % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqr1ngB % PrgifHhDYfgasaacH8srps0lbbf9q8WrFfeuY-Hhbba9q8qqaqFr0x % c9pk0xbba9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8fr % Fve9Fve9Ff0dmeaabiqaceGabiGaaeaabaWaaeaaeaaakeaacaaIWa % GaeyipaWJaamyyaiabgYda8iaaigdaaaa!3BB7! 0 < a < 1
Câu 38: Tìm số phức z thỏa mãn |z - 2| = |z| và % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WG6bGaey4kaSIaaGymaaGaayjkaiaawMcaamaabmaabaGabmOEayaa % raGaeyOeI0IaamyAaaGaayjkaiaawMcaaaaa!3E94! \left( {z + 1} \right)\left( {\bar z - i} \right)
Câu 39: Lớp 11A có 40 học sinh trong đó có 12 học sinh đạt điểm tổng kết môn Hóa học loại giỏi và 13 học sinh đạt điểm tổng kết môn Vật lí loại giỏi. Biết rằng khi chọn một học sinh của lớp đạt điểm tổng kết môn Hóa học hoặc Vật lí loại giỏi có xác suất là 0,5. Số học sinh đạt điểm tổng kết giỏi cả hai môn Hóa học và Vật lí là
Câu 40: Công thức nào sau đây là đúng với cấp số cộng có số hạng đầu u_1\) , công sai d, \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOBaiabgw % MiZkaaikdacaGGUaaaaa!3A1A! n \ge 2.
A. % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyDamaaBa % aaleaacaWGUbaabeaakiabg2da9iaadwhadaWgaaWcbaGaaGymaaqa % baGccqGHRaWkdaqadaqaaiaad6gacqGHsislcaaIXaaacaGLOaGaay % zkaaGaamizaaaa!40F6! {u_n} = {u_1} + \left( {n - 1} \right)d B. % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyDamaaBa % aaleaacaWGUbaabeaakiabg2da9iaadwhadaWgaaWcbaGaaGymaaqa % baGccqGHRaWkdaqadaqaaiaad6gacqGHRaWkcaaIXaaacaGLOaGaay % zkaaGaamizaaaa!40EB! {u_n} = {u_1} + \left( {n + 1} \right)d C. % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyDamaaBa % aaleaacaWGUbaabeaakiabg2da9iaadwhadaWgaaWcbaGaaGymaaqa % baGccqGHsisldaqadaqaaiaad6gacqGHsislcaaIXaaacaGLOaGaay % zkaaGaamizaaaa!4101! {u_n} = {u_1} - \left( {n - 1} \right)d D. % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyDamaaBa % aaleaacaWGUbaabeaakiabg2da9iaadwhadaWgaaWcbaGaaGymaaqa % baGccqGHRaWkcaWGKbaaaa!3CD2! {u_n} = {u_1} + d
Câu 41: Cho a,b,c là các số thực sao cho phương trình % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXafv3ySLgzGmvETj2BSbqefm0B1jxALjhiov2D % aerbdfgBPjMCPbqeeuuDJXwAKbsr4rNCHbGeaGqiVu0Je9sqqrpepC % 0xbbL8F4rqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yq % aqpepae9pg0FirpepeKkFr0xfr-xfr-xb9adbaqaaeGaciGaaiaabe % qaamaaeaqbaaGcbaGaamOEamaaCaaaleqabaGaaG4maaaakiabgUca % RiaadggacaWG6bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaamOyai % aadQhacqGHRaWkcaWGJbGaeyypa0JaaGimaaaa!48ED! {z^3} + a{z^2} + bz + c = 0\)</span> có ba nghiệm phức lần lượt là <span class="math-tex">\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXafv3ySLgzGmvETj2BSbqefm0B1jxALjhiov2D % aerbdfgBPjMCPbqeeuuDJXwAKbsr4rNCHbGeaGqiVu0Je9sqqrpepC % 0xbbL8F4rqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yq % aqpepae9pg0FirpepeKkFr0xfr-xfr-xb9adbaqaaeGaciGaaiaabe % qaamaaeaqbaaGcbaGaamOEamaaBaaaleaacaaIXaaabeaakiabg2da % 9iabeM8a3jabgUcaRiaaiodacaWGPbGaai4oaiaabccacaWG6bWaaS % baaSqaaiaaikdaaeqaaOGaeyypa0JaeqyYdCNaey4kaSIaaGyoaiaa % dMgacaGG7aGaaeiiaiaadQhadaWgaaWcbaGaaG4maaqabaGccqGH9a % qpcaaIYaGaeqyYdCNaeyOeI0IaaGinaaaa!5585! {z_1} = \omega + 3i;{\rm{ }}{z_2} = \omega + 9i;{\rm{ }}{z_3} = 2\omega - 4\)</span>, trong đó <span class="math-tex">\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXafv3ySLgzGmvETj2BSbqefm0B1jxALjhiov2D % aerbdfgBPjMCPbqeeuuDJXwAKbsr4rNCHbGeaGqiVu0Je9sqqrpepC % 0xbbL8F4rqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yq % aqpepae9pg0FirpepeKkFr0xfr-xfr-xb9adbaqaaeGaciGaaiaabe % qaamaaeaqbaaGcbaGaeqyYdChaaa!3EBB! \omega \)</span> là một số phức nào đó. Tính giá trị của <span class="math-tex">\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXafv3ySLgzGmvETj2BSbqefm0B1jxALjhiov2D % aerbdfgBPjMCPbqeeuuDJXwAKbsr4rNCHbGeaGqiVu0Je9sqqrpepC % 0xbbL8F4rqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yq % aqpepae9pg0FirpepeKkFr0xfr-xfr-xb9adbaqaaeGaciGaaiaabe % qaamaaeaqbaaGcbaGaamiuaiabg2da9maaemaabaGaamyyaiabgUca % RiaadkgacqGHRaWkcaWGJbaacaGLhWUaayjcSdGaaiOlaaaa!4716! P = \left| {a + b + c} \right|.
Câu 42: Cho hàm số y = f(x). Khẳng định nào sau đây là đúng ?
A. Hàm số y = f(x) đạt cực trị tại
B. Hàm số y = f(x) đạt cực trị tại
% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEamaaBa % aaleaacaaIWaaabeaaaaa!37D7! {x_0}\)</span> thì <span class="math-tex">\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafa % WaaeWaaeaacaWG4bWaaSbaaSqaaiaaicdaaeqaaaGccaGLOaGaayzk % aaGaeyypa0JaaGimaaaa!3C21! f'\left( {{x_0}} \right) = 0 .
C. Hàm số y = f(x) đạt cực trị tại
% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEamaaBa % aaleaacaaIWaaabeaaaaa!37D7! {x_0}\)</span> thì nó không có đạo hàm tại <span class="math-tex">\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEamaaBa % aaleaacaaIWaaabeaaaaa!37D7! {x_0} .
D. Nếu hàm số đạt cực trị tại
% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEamaaBa % aaleaacaaIWaaabeaaaaa!37D7! {x_0}\)</span> thì hàm số không có đạo hàm tại <span class="math-tex">\({x_0}\)</span> hoặc <span class="math-tex">\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafa % WaaeWaaeaacaWG4bWaaSbaaSqaaiaaicdaaeqaaaGccaGLOaGaayzk % aaGaeyypa0JaaGimaaaa!3C21! f'\left( {{x_0}} \right) = 0 .
Câu 43: Cho A(1;-3;2) và mặt phẳng % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaqadaqaaiaadcfaaiaawIcacaGLPaaacaGG6aGaaGOmaiaadIha % cqGHsislcaWG5bGaey4kaSIaaG4maiaadQhacqGHsislcaaIXaGaey % ypa0JaaGimaaaa!42DA! \left( P \right):2x - y + 3z - 1 = 0
A. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaGabaabaeqabaGaamiEaiabg2da9iaaikdacqGHRaWkcaWG0baa % baGaamyEaiabg2da9iabgkHiTiaaigdacqGHsislcaaIZaGaamiDaa % qaaiaadQhacqGH9aqpcaaIZaGaey4kaSIaaGOmaiaadshaaaGaay5E % aaaaaa!4778! \left\{ \begin{array}{l} x = 2 + t\\ y = - 1 - 3t\\ z = 3 + 2t \end{array} \right. B. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaGabaabaeqabaGaamiEaiabg2da9iaaigdacqGHRaWkcaaIYaGa % amiDaaqaaiaadMhacqGH9aqpcqGHsislcaaIZaGaey4kaSIaamiDaa % qaaiaadQhacqGH9aqpcaaIYaGaey4kaSIaaG4maiaadshaaaGaay5E % aaaaaa!476D! \left\{ \begin{array}{l} x = 1 + 2t\\ y = - 3 + t\\ z = 2 + 3t \end{array} \right. C. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaGabaabaeqabaGaamiEaiabg2da9iaaigdacqGHRaWkcaaIYaGa % amiDaaqaaiaadMhacqGH9aqpcqGHsislcaaIZaGaeyOeI0IaamiDaa % qaaiaadQhacqGH9aqpcaaIYaGaey4kaSIaaG4maiaadshaaaGaay5E % aaaaaa!4778! \left\{ \begin{array}{l} x = 1 + 2t\\ y = - 3 - t\\ z = 2 + 3t \end{array} \right. D. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaGabaabaeqabaGaamiEaiabg2da9iaaigdacqGHRaWkcaaIYaGa % amiDaaqaaiaadMhacqGH9aqpcqGHsislcaaIZaGaeyOeI0IaamiDaa % qaaiaadQhacqGH9aqpcaaIYaGaeyOeI0IaaG4maiaadshaaaGaay5E % aaaaaa!4783! \left\{ \begin{array}{l} x = 1 + 2t\\ y = - 3 - t\\ z = 2 - 3t \end{array} \right.
Câu 44: Trong không gian với hệ tọa độ Oxyz, cho hai điểm A( -3;1; -4) và B(1; -1;2). Phương trình mặt cầu (S) nhận AB làm đường kính là
A. % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WG4bGaey4kaSIaaGymaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOm % aaaakiabgUcaRiaadMhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkda % qadaqaaiaadQhacqGHRaWkcaaIXaaacaGLOaGaayzkaaWaaWbaaSqa % beaacaaIYaaaaOGaeyypa0JaaGynaiaaiAdaaaa!465B! {\left( {x + 1} \right)^2} + {y^2} + {\left( {z + 1} \right)^2} = 56 B. % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WG4bGaeyOeI0IaaGinaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOm % aaaakiabgUcaRmaabmaabaGaamyEaiabgUcaRiaaikdaaiaawIcaca % GLPaaadaahaaWcbeqaaiaaikdaaaGccqGHRaWkdaqadaqaaiaadQha % cqGHsislcaaI2aaacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaO % Gaeyypa0JaaGymaiaaisdaaaa!499A! {\left( {x - 4} \right)^2} + {\left( {y + 2} \right)^2} + {\left( {z - 6} \right)^2} = 14 C. % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WG4bGaey4kaSIaaGymaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOm % aaaakiabgUcaRiaadMhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkda % qadaqaaiaadQhacqGHRaWkcaaIXaaacaGLOaGaayzkaaWaaWbaaSqa % beaacaaIYaaaaOGaeyypa0JaaGymaiaaisdaaaa!4655! {\left( {x + 1} \right)^2} + {y^2} + {\left( {z + 1} \right)^2} = 14 D. % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WG4bGaeyOeI0IaaGymaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOm % aaaakiabgUcaRiaadMhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkda % qadaqaaiaadQhacqGHsislcaaIXaaacaGLOaGaayzkaaWaaWbaaSqa % beaacaaIYaaaaOGaeyypa0JaaGymaiaaisdaaaa!466B! {\left( {x - 1} \right)^2} + {y^2} + {\left( {z - 1} \right)^2} = 14
Câu 45: Cho tứ diện ABCD có AB = 3a,AC = 4a ,AD = 5a. Gọi M,N,P lần lượt là trọng tâm các tam giác DAB ,DBC ,DCA . Tính thể tích V của tứ diện DMDMNP khi thể tích tứ diện ABCD đạt giá trị lớn nhất.
A. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXafv3ySLgzGmvETj2BSbqefm0B1jxALjhiov2D % aebbfv3ySLgzGueE0jxyaibaieYlh9qrpeeu0dXdh9vqqj-hEeeu0x % Xdbba9frFj0-OqFfea0dXdd9vqaq-JfrVkFHe9pgea0dXdar-Jb9hs % 0dXdbPYxe9vr0-vr0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaO % qaaiaadAfacqGH9aqpdaWcaaqaaiaaigdacaaIYaGaaGimaiaadgga % daahaaWcbeqaaiaaiodaaaaakeaacaaIYaGaaG4naaaaaaa!40FB! V = \frac{{120{a^3}}}{{27}} B. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXafv3ySLgzGmvETj2BSbqefm0B1jxALjhiov2D % aebbfv3ySLgzGueE0jxyaibaieYlh9qrpeeu0dXdh9vqqj-hEeeu0x % Xdbba9frFj0-OqFfea0dXdd9vqaq-JfrVkFHe9pgea0dXdar-Jb9hs % 0dXdbPYxe9vr0-vr0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaO % qaaiaadAfacqGH9aqpdaWcaaqaaiaaigdacaaIWaGaamyyamaaCaaa % leqabaGaaG4maaaaaOqaaiaaisdaaaaaaa!3F80! V = \frac{{10{a^3}}}{4} C. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXafv3ySLgzGmvETj2BSbqefm0B1jxALjhiov2D % aebbfv3ySLgzGueE0jxyaibaieYlh9qrpeeu0dXdh9vqqj-hEeeu0x % Xdbba9frFj0-OqFfea0dXdd9vqaq-JfrVkFHe9pgea0dXdar-Jb9hs % 0dXdbPYxe9vr0-vr0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaO % qaaiaadAfacqGH9aqpdaWcaaqaaiaaiIdacaaIWaGaamyyamaaCaaa % leqabaGaaG4maaaaaOqaaiaaiEdaaaaaaa!3F8A! V = \frac{{80{a^3}}}{7} D. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXafv3ySLgzGmvETj2BSbqefm0B1jxALjhiov2D % aebbfv3ySLgzGueE0jxyaibaieYlh9qrpeeu0dXdh9vqqj-hEeeu0x % Xdbba9frFj0-OqFfea0dXdd9vqaq-JfrVkFHe9pgea0dXdar-Jb9hs % 0dXdbPYxe9vr0-vr0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaO % qaaiaadAfacqGH9aqpdaWcaaqaaiaaikdacaaIWaGaamyyamaaCaaa % leqabaGaaG4maaaaaOqaaiaaikdacaaI3aaaaaaa!4040! V = \frac{{20{a^3}}}{{27}}
Câu 46: Cho hai điểm A(3;3;1),B(0;2;1), mặt phẳng % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WGqbaacaGLOaGaayzkaaGaaiOoaiaadIhacqGHRaWkcaWG5bGaey4k % aSIaamOEaiabgkHiTiaaiEdacqGH9aqpcaaIWaaaaa!413C! \left( P \right):x + y + z - 7 = 0
A. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadIhacqGH9aqpcaWG0baabaGaamyEaiabg2da9iaaiEdacqGH % sislcaaIZaGaamiDaaqaaiaadQhacqGH9aqpcaaIYaGaamiDaaaaca % GL7baaaaa!4334! \left\{ \begin{array}{l} x = t\\ y = 7 - 3t\\ z = 2t \end{array} \right. B. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadIhacqGH9aqpcqGHsislcaWG0baabaGaamyEaiabg2da9iaa % iEdacqGHsislcaaIZaGaamiDaaqaaiaadQhacqGH9aqpcaaIYaGaam % iDaaaacaGL7baaaaa!4421! \left\{ \begin{array}{l} x = - t\\ y = 7 - 3t\\ z = 2t \end{array} \right. C. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadIhacqGH9aqpcaWG0baabaGaamyEaiabg2da9iaaiEdacqGH % RaWkcaaIZaGaamiDaaqaaiaadQhacqGH9aqpcaaIYaGaamiDaaaaca % GL7baaaaa!4329! \left\{ \begin{array}{l} x = t\\ y = 7 + 3t\\ z = 2t \end{array} \right. D. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadIhacqGH9aqpcaaIYaGaamiDaaqaaiaadMhacqGH9aqpcaaI % 3aGaeyOeI0IaaG4maiaadshaaeaacaWG6bGaeyypa0JaaGOmaiaads % haaaGaay5Eaaaaaa!43F0! \left\{ \begin{array}{l} x = 2t\\ y = 7 - 3t\\ z = 2t \end{array} \right.
Câu 47: Tổng số đỉnh, số cạnh và số mặt của hình lập phương là
Câu 48: Tập xác định của hàm số % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maabmaabaGaaGOmaiabgkHiTiaadIhaaiaawIcacaGLPaaadaah % aaWcbeqaamaakaaabaGaaG4maaadbeaaaaaaaa!3D2D! y = {\left( {2 - x} \right)^{\sqrt 3 }}
A. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiraiabg2 % da9maabmaabaGaaGOmaiaacUdacqGHRaWkcqGHEisPaiaawIcacaGL % Paaaaaa!3D1A! D = \left( {2; + \infty } \right) B. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiraiabg2 % da9maabmaabaGaeyOeI0IaeyOhIuQaai4oaiaaikdaaiaawIcacaGL % Paaaaaa!3D25! D = \left( { - \infty ;2} \right) C. % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiraiabg2 % da9maajadabaGaeyOeI0IaeyOhIuQaai4oaiaaikdaaiaawIcacaGL % Dbaaaaa!3D8E! D = \left( { - \infty ;2} \right]
Câu 49: Đồ thị (C) của hàm số % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqabeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maalaaabaGaamiEaiabgUcaRiaaigdaaeaacaWG4bGaeyOeI0Ia % aGymaaaaaaa!3D48! y = \frac{{x + 1}}{{x - 1}}
A. % MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGOmamaaka % aabaGaaG4maaWcbeaaaaa!3787! 2\sqrt 3
Câu 50: Cho hàm số % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaadggacaWG4bWaaWbaaSqabeaacaaIZaaaaOGaey4kaSIaamOy % aiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaWGJbGaamiEai % abgUcaRiaaigdaaaa!42EC! y = a{x^3} + b{x^2} + cx + 1
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