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

a) $\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}=\sqrt{2^2.5}-\sqrt{3^2.5}+3\sqrt{3^2.2}+\sqrt{6^2.2}$

$=2\sqrt{5}-3\sqrt{5}+9\sqrt{2}+6\sqrt{2}=\left(2-3\right)\sqrt{5}+\left(9+6\right)\sqrt{2}=15\sqrt{2}-\sqrt{5}$

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

$=\left(\sqrt{2^2.7}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{2^2.21}$

$$\begin{gathered} =2\sqrt{7}.\sqrt{7}-2\sqrt{3}.\sqrt{7}+\sqrt{7}.\sqrt{7}+2\sqrt{21} \\ =2.7-2\sqrt{21}+7+2\sqrt{21} \\ =14+7 \\ =21 \end{gathered}$$

a) $\mathrm{M}=\left|\sqrt{5}-1\right|-\sqrt{5}=\sqrt{5}-1-\sqrt{5}=-1$

b) $\mathrm{N}=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}=\sqrt{5+2\sqrt{5}+1}-\sqrt{5-2\sqrt{5}+1}=\sqrt{{\left(\sqrt{5}+1\right)}^2}-\sqrt{{\left(\sqrt{5}-1\right)}^2}$

$=\left|\sqrt{5}+1\right|-\left|\sqrt{5}-1\right|=\sqrt{5}+1-\sqrt{5}+1=2$

a) $\mathrm{A}=\sqrt{7-2\sqrt{10}}+\sqrt{20}+\frac{1}{2}\sqrt{8}=\sqrt{{\left(\sqrt{5}-\sqrt{2}\right)}^2}+2\sqrt{5}+\frac{1}{2}.2\sqrt{2}$

$=\left|\sqrt{5}-\sqrt{2}\right|+2\sqrt{5}+\sqrt{2}=\sqrt{5}-\sqrt{2}+2\sqrt{5}+\sqrt{2}=3\sqrt{5}$

b) $\mathrm{B}=\frac{21}{2}{\left(\sqrt{4+2\sqrt{3}}+\sqrt{6-2\sqrt{5}}\right)}^2-3{\left(\sqrt{4-2\sqrt{3}}+\sqrt{6+2\sqrt{5}}\right)}^2-15\sqrt{15}$

$=\frac{21}{2}{\left(\sqrt{3}+1+\sqrt{5}-1\right)}^2-3{\left(\sqrt{3}-1+\sqrt{5}+1\right)}^2-15\sqrt{15}=\frac{15}{2}{\left(\sqrt{3}+\sqrt{5}\right)}^2-15\sqrt{15}=60$

$$\begin{gathered} \mathrm{a}) \mathrm{Ta} \mathrm{có}: \mathrm{P}=\frac{\sqrt{5}+2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}+\frac{\sqrt{5}-2}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)} \\ =\sqrt{5}+2+\sqrt{5}-2=2\sqrt{5} \\ \mathrm{Vậy} \mathrm{P}=2\sqrt{5} \end{gathered}$$

$$\begin{gathered} \mathrm{b}) \mathrm{Q}=\frac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}-\frac{2+\sqrt{8}}{1+\sqrt{2}} \\ =\frac{\sqrt{3}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\frac{2\left(1+\sqrt{2}\right)}{1+\sqrt{2}} \\ =\sqrt{3}-2 \\ \mathrm{Vậy} \mathrm{Q}=\sqrt{3}-2 \end{gathered}$$

a) Ta có: $\mathrm{A}=\frac{1}{\sqrt{3}+\sqrt{2}}-\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}$

$$\begin{gathered} =\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}-\frac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2} \\ =\sqrt{3}-\sqrt{2}-\sqrt{3}=-\sqrt{2} \\ \mathrm{Vậy} \mathrm{A}=-\sqrt{2} \end{gathered}$$

$$\begin{gathered} \mathrm{b}) \mathrm{B}=\frac{\sqrt{3}-1+\sqrt{3}+1}{\sqrt{3}+1}++\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{\sqrt{2}} \\ =\frac{2\sqrt{3}}{3-1}+2-\sqrt{3} \\ =\sqrt{3}+2-\sqrt{3} \\ =2 \\ \mathrm{Vậy} \mathrm{B}=2 \end{gathered}$$

a) $\mathrm{A}=\frac{1}{\sqrt{2}+1}-\frac{\sqrt{8}-\sqrt{10}}{2-\sqrt{5}}=\frac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}-\frac{\sqrt{2}\left(2-\sqrt{5}\right)}{2-\sqrt{5}}=\sqrt{2}-1-\sqrt{2}=-1$

Vậy A = -1

b) $\mathrm{B}=\frac{2}{\sqrt{7}-\sqrt{6}}-\sqrt{28}+\sqrt{54}=\frac{2\left(\sqrt{7}+\sqrt{6}\right)}{\left(\sqrt{7}-\sqrt{6}\right)\left(\sqrt{7}+\sqrt{6}\right)}-\sqrt{7.4}+\sqrt{9.6}=\frac{2\sqrt{7}+2\sqrt{6}}{7-6}-2\sqrt{7}+3\sqrt{6}$

$=2\sqrt{7}+2\sqrt{6}-2\sqrt{7}+3\sqrt{6}=5\sqrt{6}$

$\mathrm{Vậy} \mathrm{B}=5\sqrt{6}$

c) $$\begin{gathered} \mathrm{C}=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-2\sqrt{3}+\sqrt{3}=\frac{2\left(\sqrt{3}+1\right)}{3-1}-2\sqrt{3}=\sqrt{3}+1-2\sqrt{3}=1-\sqrt{3} \\ \mathrm{Vậy} \mathrm{C}=1-\sqrt{3} \end{gathered}$$

$$\begin{gathered} \mathrm{d}) \mathrm{D}=\frac{5+\sqrt{5}}{\sqrt{5}+2}+\frac{\sqrt{5}}{\sqrt{5}-1}-\frac{3\sqrt{5}}{3+\sqrt{5}}=\frac{\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}+\frac{\sqrt{5}\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}-\frac{3\sqrt{5}\left(3-\sqrt{5}\right)}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)} \\ =3\sqrt{5}-5+\frac{{\displaystyle 5+\sqrt{5}}}{{\displaystyle 4}}-\frac{{\displaystyle 9\sqrt{5}-15}}{{\displaystyle 4}}=3\sqrt{5}-5+\frac{{\displaystyle 5+\sqrt{5}-9\sqrt{5}+15}}{{\displaystyle 4}}=3\sqrt{5}-5+5-2\sqrt{5}=\sqrt{5} \\ \mathrm{Vậy} \mathrm{D}=\sqrt{5} \end{gathered}$$

a) Ta có: $$\begin{gathered} \mathrm{A}=\frac{\sqrt{2}}{1+\sqrt{2}}+\frac{\sqrt{9.2}}{3}=\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{2-1}+\frac{3\sqrt{2}}{3} \\ =\frac{2-\sqrt{2}}{1}+\sqrt{2}=2-\sqrt{2}+\sqrt{2}=2 \end{gathered}$$

$$\begin{gathered} \mathrm{b}) \mathrm{B}=\frac{3\left(\sqrt{7}+2\right)}{\left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right)}-\frac{2.{\left(\sqrt{7}\right)}^2}{\sqrt{7}}+\left|\sqrt{7}-2\right| \\ =\frac{3\left(\sqrt{7}+2\right)}{3}-2\sqrt{7}+\sqrt{7}-2=\sqrt{7}+2-\sqrt{7}-2=0 \end{gathered}$$

Vậy B = 0

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

Vậy C = 2

a) $\mathrm{A}={\left(\sqrt{5}\right)}^2-2^2-\frac{\sqrt{{\left(2-\sqrt{3}\right)}^2}}{\sqrt{3}-2}=5-4-\frac{2-\sqrt{3}}{\sqrt{3}-2}=1-\left(-1\right)=2$

Vậy A = 2

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

$$\begin{gathered} =\frac{{\left(2+\sqrt{3}\right)}^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}-\frac{{\left(2-\sqrt{3}\right)}^2}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)} \\ ={\left(2+\sqrt{3}\right)}^2-{\left(2-\sqrt{3}\right)}^2 \\ =\left(2+\sqrt{3}+2-\sqrt{3}\right)\left(2+\sqrt{3}-2+\sqrt{3}\right) \\ =4.2\sqrt{3}=8\sqrt{3} \\ \mathrm{Vậy} \mathrm{B}=8\sqrt{3} \end{gathered}$$

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

$=\sqrt{\frac{22-11\sqrt{3}}{11}}-\sqrt{\frac{26+13\sqrt{3}}{13}}=\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}=\sqrt{\frac{4-2\sqrt{3}}{2}}-\sqrt{\frac{4+2\sqrt{3}}{2}}$

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

Vậy $\mathrm{C}=-\sqrt{2}$

a) $\mathrm{A}=\left(3\sqrt{50}-5\sqrt{18}+3\sqrt{8}\right)\sqrt{2}=\left(15\sqrt{2}-15\sqrt{2}+6\sqrt{2}\right)\sqrt{2}=6\sqrt{2}.\sqrt{2}=12$

b) $\mathrm{B}=\sqrt{{\left(\sqrt{3}-1\right)}^2}-\sqrt{3}=\sqrt{3}-1-\sqrt{3}=-1$

c) $\mathrm{C}={\left(\sqrt{5}-\sqrt{2}\right)}^2+\sqrt{40}=5-2\sqrt{10}+2+2\sqrt{10}=7$

d) $D=5\sqrt{8}+\sqrt{50}-2\sqrt{18}=10\sqrt{2}+5\sqrt{2}-6\sqrt{2}=\left(10+5-6\right)\sqrt{2}=9\sqrt{2}$

e) $\mathrm{E}=2\sqrt{32}-5\sqrt{27}-4\sqrt{8}+3\sqrt{75}=8\sqrt{2}-15\sqrt{3}-8\sqrt{2}+15\sqrt{3}=0$

a) A = $5\sqrt{5}-12\sqrt{5}+6\sqrt{5}-4\sqrt{5}=-5\sqrt{5}$

b) $\mathrm{B}=\left(3\sqrt{2}+\sqrt{6}\right)\sqrt{6-3\sqrt{3}}=\left(3+\sqrt{3}\right)\sqrt{12-6\sqrt{3}}$

$=\left(3+\sqrt{3}\right)\left|3-\sqrt{3}\right|=\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)=9-3=6$

a) $A=\frac{1}{3+\sqrt{7}}+\frac{1}{3-\sqrt{7}}=\frac{6}{3^2-{\sqrt{7}}^2}=\frac{6}{9-7}=3$

b) $B=\left|\sqrt{3}+2\right|+\left|\sqrt{3}-2\right|=\sqrt{3}+2-\sqrt{3}+2=4$

$$\begin{gathered} \mathrm{c}) C=\left(2+\frac{3+\sqrt{3}}{\sqrt{3}+1}\right).\left(2-\frac{3-\sqrt{3}}{\sqrt{3}-1}\right)=\left[2+\frac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}\right]\left[2-\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right] \\ =\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)=1 \end{gathered}$$

$$\begin{gathered} \mathrm{d}) D=\left(2+\frac{5-\sqrt{5}}{\sqrt{5}-1}\right).\left(2-\frac{5+\sqrt{5}}{\sqrt{5}+1}\right)=\left[2+\frac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right]\left[2-\frac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\right] \\ =\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)=-1 \end{gathered}$$

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

$=\left(\sqrt{3}+1\right)\sqrt{\frac{88-44\sqrt{3}}{22}}=\left(\sqrt{3}+1\right)\sqrt{4-2\sqrt{3}}=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)=2$

Cách 2: $E=\left(\sqrt{3}+1\right)\sqrt{\frac{14-6\sqrt{3}}{5+\sqrt{3}}}=\sqrt{\frac{\left(4+2\sqrt{3}\right)\left(14-6\sqrt{3}\right)}{5+\sqrt{3}}}=\sqrt{\frac{20+4\sqrt{3}}{5+\sqrt{3}}}=\sqrt{4}=2$