Đáp án:
$\begin{array}{l}
62)Dkxd:\left\{ \begin{array}{l}
2x + 2 \ge 0\\
x + 1 > 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \ge - 1\\
x > - 1
\end{array} \right. \Rightarrow x > - 1\\
x\sqrt {2x + 2} + \left( {x + 1} \right).\sqrt {\dfrac{2}{{x + 1}}} - 4\sqrt {\dfrac{{x + 1}}{2}} \\
= x\sqrt {2\left( {x + 1} \right)} + {\left( {\sqrt {x + 1} } \right)^2}.\dfrac{{\sqrt 2 }}{{\sqrt {x + 1} }} - 4.\dfrac{{\sqrt {x + 1} }}{{\sqrt 2 }}\\
= \sqrt 2 .x\sqrt {x + 1} + \sqrt 2 .\sqrt {x + 1} - 2\sqrt 2 .\sqrt {x + 1} \\
= \sqrt 2 .x\sqrt {x + 1} - \sqrt 2 .\sqrt {x + 1} \\
= \sqrt 2 .\sqrt {x + 1} .\left( {x - 1} \right)\\
64)\\
Dkxd:x \ge 0;x \ne 9\\
\dfrac{{\sqrt x + 2}}{{\sqrt x - 3}} - \dfrac{{5\sqrt x }}{{\sqrt x + 3}} = \dfrac{{22}}{{x - 9}}\\
\Leftrightarrow \dfrac{{\sqrt x + 2}}{{\sqrt x - 3}} - \dfrac{{5\sqrt x }}{{\sqrt x + 3}} = \dfrac{{22}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
\Leftrightarrow \dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x + 3} \right) - 5\sqrt x \left( {\sqrt x - 3} \right) - 22}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}} = 0\\
\Rightarrow x + 5\sqrt x + 6 - 5x + 15\sqrt x - 22 = 0\\
\Rightarrow - 4x + 20\sqrt x - 16 = 0\\
\Rightarrow x - 5\sqrt x + 4 = 0\\
\Rightarrow \left( {\sqrt x - 1} \right)\left( {\sqrt x - 4} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\sqrt x = 1\\
\sqrt x = 4
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 1\left( {tmdk} \right)\\
x = 16\left( {tmdk} \right)
\end{array} \right.\\
\text{Vậy}\,x = 1;x = 16
\end{array}$
