Đáp án:

\(\begin{array}{l}
B4:\\
a)\dfrac{{\sqrt {10} }}{2}\\
b)\dfrac{{\sqrt 5 }}{2}\\
c)\dfrac{{\sqrt {20} }}{{60}}\\
d)\dfrac{{4 + 2\sqrt 2 }}{{10}}\\
e)\dfrac{{3\sqrt 3  - 3}}{2}\\
f)3 + 2\sqrt 3 \\
g)7 + 4\sqrt 3 \\
h)\sqrt {10}  - \sqrt 7 \\
j)\sqrt 2 \\
k) - \sqrt 5 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
B2:\\
a)\dfrac{2}{{x - 3}}.\sqrt {\dfrac{{{{\left( {x - 3} \right)}^2}}}{{4{y^4}}}} \\
 = \dfrac{2}{{x - 3}}.\dfrac{{ - \left( {x - 3} \right)}}{{2{y^2}}}\\
 =  - \dfrac{1}{{{y^2}}}\\
b)\dfrac{2}{{2x - 1}}.\sqrt {5{x^2}{{\left( {1 - 2x} \right)}^2}} \\
 = \dfrac{2}{{2x - 1}}.\left[ { - x.\left( {1 - 2x} \right).\sqrt 5 } \right]\\
 = \dfrac{2}{{2x - 1}}.x\left( {2x - 1} \right).\sqrt 5 \\
 = 2x\sqrt 5 \\
B3:\\
a)\dfrac{1}{{10\sqrt 6 }} = \dfrac{{\sqrt 6 }}{{60}}\\
b)\dfrac{{\sqrt {11} }}{{6\sqrt {15} }} = \dfrac{{\sqrt {11.15} }}{{6.15}} = \dfrac{{\sqrt {165} }}{{90}}\\
c)\dfrac{{\sqrt 3 }}{{5\sqrt 2 }} = \dfrac{{\sqrt 6 }}{{10}}\\
d)\dfrac{{\sqrt 5 }}{{7\sqrt 2 }} = \dfrac{{\sqrt {10} }}{{14}}\\
e)\dfrac{{\left| {1 - \sqrt 3 } \right|}}{{3\sqrt 3 }} = \dfrac{{\sqrt 3 \left( {\sqrt 3  - 1} \right)}}{9}\\
 = \dfrac{{9 - \sqrt 3 }}{9}\\
B4:\\
a)\dfrac{{\sqrt 5 }}{{\sqrt 2 }} = \dfrac{{\sqrt {10} }}{2}\\
b)\dfrac{5}{{2\sqrt 5 }} = \dfrac{{\sqrt 5 }}{2}\\
c)\dfrac{{\sqrt {20} }}{{60}}\\
d)\dfrac{{\left( {2\sqrt 2  + 2} \right)\sqrt 2 }}{{10}}\\
 = \dfrac{{4 + 2\sqrt 2 }}{{10}}\\
e)\dfrac{{3\left( {\sqrt 3  - 1} \right)}}{{\left( {\sqrt 3  + 1} \right)\left( {\sqrt 3  - 1} \right)}} = \dfrac{{3\sqrt 3  - 3}}{{3 - 1}}\\
 = \dfrac{{3\sqrt 3  - 3}}{2}\\
f)\dfrac{{\sqrt 3 \left( {\sqrt 3  + 2} \right)}}{{\left( {\sqrt 3  + 2} \right)\left( {\sqrt 3  - 2} \right)}}\\
 = \dfrac{{3 + 2\sqrt 3 }}{{3 - 2}} = 3 + 2\sqrt 3 \\
g)\dfrac{{{{\left( {2 + \sqrt 3 } \right)}^2}}}{{\left( {2 + \sqrt 3 } \right)\left( {2 - \sqrt 3 } \right)}}\\
 = \dfrac{{7 + 4\sqrt 3 }}{{4 - 3}} = \dfrac{{7 + 4\sqrt 3 }}{1} = 7 + 4\sqrt 3 \\
h)\dfrac{{3\left( {\sqrt {10}  - \sqrt 7 } \right)}}{{\left( {\sqrt {10}  - \sqrt 7 } \right)\left( {\sqrt {10}  - \sqrt 7 } \right)}}\\
 = \dfrac{{3\left( {\sqrt {10}  - \sqrt 7 } \right)}}{{10 - 7}} = \sqrt {10}  - \sqrt 7 \\
j)\dfrac{{\sqrt 2 \left( {\sqrt 2  + 1} \right)}}{{1 + \sqrt 2 }} = \sqrt 2 \\
k)\dfrac{{\sqrt 5 \left( {\sqrt 3  - 1} \right)}}{{1 - \sqrt 3 }} =  - \sqrt 5 
\end{array}\)