Đáp án:
$\begin{array}{l}
d)\sqrt {\dfrac{2}{a}} :\sqrt {\dfrac{{50}}{{{a^3}}}} \\
= \sqrt {\dfrac{2}{a}.\dfrac{{{a^3}}}{{50}}} \\
= \sqrt {\dfrac{{{a^2}}}{{25}}} \\
= \dfrac{a}{5}\left( {do:a > 0} \right)\\
e)\dfrac{{\sqrt {{x^2} - 2x + 1} }}{{\sqrt {{x^2} + 2x + 1} }}\\
= \dfrac{{\sqrt {{{\left( {x - 1} \right)}^2}} }}{{\sqrt {{{\left( {x + 1} \right)}^2}} }}\\
= \dfrac{{x - 1}}{{x + 1}}\left( {do:x > 1} \right)\\
f)\sqrt {\dfrac{{{x^2} - 4x + 4}}{{9 - 6x + {x^2}}}} + \dfrac{{{x^2} - 1}}{{x - 3}}\\
= \sqrt {\dfrac{{{{\left( {x - 2} \right)}^2}}}{{{{\left( {3 - x} \right)}^2}}}} + \dfrac{{{x^2} - 1}}{{x - 3}}\\
= \left[ \begin{array}{l}
\dfrac{{x - 2}}{{3 - x}} + \dfrac{{{x^2} - 1}}{{x - 3}}\left( {khi:2 \le x < 3} \right)\\
\dfrac{{2 - x}}{{3 - x}} + \dfrac{{{x^2} - 1}}{{x - 3}}\left( {khi:x < 2} \right)
\end{array} \right.\\
= \left[ \begin{array}{l}
\dfrac{{{x^2} - x + 1}}{{x - 3}}\left( {khi:2 \le x < 3} \right)\\
\dfrac{{{x^2} + x - 3}}{{x - 3}}\left( {khi:x < 2} \right)
\end{array} \right.
\end{array}$
