Giải thích các bước giải:
$\begin{array}{l}
a)\sqrt {4 - 2\sqrt 3 } + \sqrt {{{\left( {\sqrt 3 - 2} \right)}^2}} \\
= \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} + \sqrt {{{\left( {\sqrt 3 - 2} \right)}^2}} \\
= \left| {\sqrt 3 - 1} \right| + \left| {\sqrt 3 - 2} \right|\\
= \sqrt 3 - 1 + 2 - \sqrt 3 \\
= 1\\
b)\sqrt {{{\left( {\sqrt 3 - 2} \right)}^2}} - \sqrt 3 \\
= \left| {\sqrt 3 - 2} \right| - \sqrt 3 \\
= 2 - \sqrt 3 - \sqrt 3 \\
= 2 - 2\sqrt 3 \\
c)\dfrac{{2 - \sqrt x }}{{4 - x}}\left( {DK:x \ge 0;x \ne 4} \right)\\
= \dfrac{{2 - \sqrt x }}{{\left( {2 - \sqrt x } \right)\left( {2 + \sqrt x } \right)}}\\
= \dfrac{1}{{2 + \sqrt x }}\\
d)\dfrac{{x - 2\sqrt x + 1}}{{\sqrt x - 1}}\left( {DK:x > 0;x \ne 1} \right)\\
= \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x - 1}}\\
= \sqrt x - 1\\
e)\dfrac{{\sqrt {15} - \sqrt 6 }}{{\sqrt {35} - \sqrt {14} }}\\
= \dfrac{{\sqrt 3 \left( {\sqrt 5 - \sqrt 2 } \right)}}{{\sqrt 7 \left( {\sqrt 5 - \sqrt 2 } \right)}}\\
= \sqrt {\dfrac{3}{7}} \\
f)\dfrac{{\sqrt {10} + \sqrt {15} }}{{\sqrt 8 + \sqrt {12} }}\\
= \dfrac{{\sqrt 5 \left( {\sqrt 2 + \sqrt 3 } \right)}}{{\sqrt 4 \left( {\sqrt 2 + \sqrt 3 } \right)}}\\
= \dfrac{{\sqrt 5 }}{2}
\end{array}$
