Đáp án:
$\begin{array}{l}
a)\sqrt {\frac{8}{{{{\left( {1 - \sqrt 2 } \right)}^2}}}} = \frac{{2\sqrt 2 }}{{\left| {1 - \sqrt 2 } \right|}}\\
= \frac{{2\sqrt 2 }}{{\sqrt 2 - 1}} = \frac{{2\sqrt 2 \left( {\sqrt 2 + 1} \right)}}{{2 - 1}}\\
= 2\sqrt 2 \left( {\sqrt 2 + 1} \right)\\
= 4 + 2\sqrt 2 \\
b)\sqrt {{{\left( {1 - x} \right)}^3}} = \sqrt {{{\left( {1 - x} \right)}^2}.\left( {1 - x} \right)} = \left( {1 - x} \right).\sqrt {1 - x} \\
\sqrt {{x^3}.{{\left( {1 - \sqrt 3 } \right)}^3}} = x\sqrt x .\left( {1 - \sqrt 3 } \right).\sqrt {\sqrt 3 - 1} \\
= \left( {1 - \sqrt 3 } \right).x.\sqrt {\left( {\sqrt 3 - 1} \right).x} \\
c)\sqrt {50.{{\left( {5 + a} \right)}^5}} = 5\sqrt 2 .{\left( {5 + a} \right)^2}.\sqrt {5 + a} \\
\sqrt {{{\left( {x - 4} \right)}^3}{{\left( {1 - x} \right)}^5}} = \left( {x - 4} \right).{\left( {1 - x} \right)^2}.\sqrt {\left( {x - 4} \right).\left( {1 - x} \right)}
\end{array}$
