Trắc nghiệm Chuyên đề toán 9 Chuyên đề 1: Biểu thức số có đáp án

b) $B=\sqrt{8}-\sqrt{18}+2\sqrt{32}$

c) $\sqrt{5}\left(\sqrt{20}-\sqrt{5}\right)+1$

b) $\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}$

c) ${\left(\sqrt{6}+\sqrt{5}\right)}^2-\sqrt{120}$

d) $\left(\frac{1}{2}\sqrt{\frac{1}{2}}-\frac{3}{2}\sqrt{2}+\frac{4}{5}\sqrt{200}\right):\frac{1}{8}$

b) $\mathrm{N}=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}$

b) $\mathrm{B}=21{\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}\right)}^2-6{\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}\right)}^2-15\sqrt{15}$

b) $\mathrm{Q}=\frac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}-\frac{2+\sqrt{8}}{1+\sqrt{2}}$

b) $\mathrm{B}=\frac{1}{\sqrt{3}+1}+\frac{1}{\sqrt{3}-1}+\frac{2\sqrt{2}-\sqrt{6}}{\sqrt{2}}$

b) $\mathrm{B}=\frac{2}{\sqrt{7}-\sqrt{6}}-\sqrt{28}+\sqrt{54}$

c) $\mathrm{C}=\frac{2}{\sqrt{3}-1}-\sqrt{27}+\frac{3}{\sqrt{3}}$

d) $\mathrm{D}=\frac{5+\sqrt{5}}{\sqrt{5}+2}+\frac{\sqrt{5}}{\sqrt{5}-1}-\frac{3\sqrt{5}}{3+\sqrt{5}}$

$\mathrm{b}) \mathrm{B}=\frac{3}{\sqrt{7}-2}-\frac{14}{\sqrt{7}}+\sqrt{{\left(\sqrt{7}-2\right)}^2}$

c) $\mathrm{C}=\left(\frac{\sqrt{21}-\sqrt{7}}{\sqrt{3}-1}+\frac{\sqrt{10}-\sqrt{5}}{\sqrt{2}-1}\right):\frac{1}{\sqrt{7}-\sqrt{5}}$

b) $\mathrm{B}=\frac{2+\sqrt{3}}{\sqrt{7-4\sqrt{3}}}-\frac{2-\sqrt{3}}{\sqrt{7+4\sqrt{3}}}$

c) $\mathrm{C}=\sqrt{\frac{3\sqrt{3}-4}{2\sqrt{3}+1}}-\sqrt{\frac{\sqrt{3}+4}{5-2\sqrt{3}}}$

a) $\mathrm{A}=\left(3\sqrt{50}-5\sqrt{18}+3\sqrt{8}\right)\sqrt{2}$

b) $\mathrm{B}=\sqrt{4-2\sqrt{3}}-\frac{1}{2}\sqrt{12}$

c) $\mathrm{C}={\left(\sqrt{5}-\sqrt{2}\right)}^2+\sqrt{40}$

d) $\mathrm{D}=5\sqrt{8}+\sqrt{50}-2\sqrt{18}$

e) $\mathrm{E}=2\sqrt{32}-5\sqrt{27}-4\sqrt{8}+3\sqrt{75}$

b) $\mathrm{B}=\left(3\sqrt{2}+\sqrt{6}\right)\sqrt{6-3\sqrt{3}}$

b) $B=\sqrt{{\left(\sqrt{3}+2\right)}^2}+\sqrt{{\left(\sqrt{3}-2\right)}^2}$

c) $C=\left(2+\frac{3+\sqrt{3}}{\sqrt{3}+1}\right).\left(2-\frac{3-\sqrt{3}}{\sqrt{3}-1}\right)$

d) $D=\left(2+\frac{5-\sqrt{5}}{\sqrt{5}-1}\right).\left(2-\frac{5+\sqrt{5}}{\sqrt{5}+1}\right)$

e) $E=\left(\sqrt{3}+1\right)\sqrt{\frac{14-6\sqrt{3}}{5+\sqrt{3}}}$