Đáp án:
$\begin{array}{l}
1)\\
{x^2} + 5 = 2\sqrt {2x + 3} - 4x\\
\Rightarrow {x^2} + 4x + 5 = 2\sqrt {2x + 3} \left( {dkxd:2x + 3 \ge 0 \Rightarrow x \ge - \frac{3}{2}} \right)\\
\Rightarrow {x^2} + 4x + 5 + 2x + 3 + 1 = 2x + 3 + 2\sqrt {2x + 3} + 1\\
\Rightarrow {x^2} + 6x + 9 = \left( {2x + 3} \right) + 2\sqrt {2x + 3} + 1\\
\Rightarrow {\left( {x + 3} \right)^2} = {\left( {\sqrt {2x + 3} + 1} \right)^2}\\
\Rightarrow \left[ \begin{array}{l}
x + 3 = \sqrt {2x + 3} + 1\\
- x - 3 = \sqrt {2x + 3} + 1\left( {loai} \right)
\end{array} \right.\\
\Rightarrow x + 2 = \sqrt {2x + 3} \\
\Rightarrow {x^2} + 4x + 4 = 2x + 3\\
\Rightarrow {x^2} + 2x + 1 = 0\\
\Rightarrow x = - 1\left( {tm} \right)\\
Vay\,x = - 1\\
2)\\
2 - {x^2} = \sqrt {2 - x} \\
\Rightarrow {\left( {2 - {x^2}} \right)^2} = 2 - x\\
\left( {dkxd:\left\{ \begin{array}{l}
2 - x \ge 0\\
2 - {x^2} \ge 0
\end{array} \right. \Rightarrow - \sqrt 2 \le x \le \sqrt 2 } \right)\\
\Rightarrow {x^4} - 4{x^2} + 4 = 2 - x\\
\Rightarrow {x^4} - 4{x^2} + x + 2 = 0\\
\Rightarrow {x^2}\left( {{x^2} - 4} \right) + x + 2 = 0\\
\Rightarrow {x^2}\left( {x + 2} \right)\left( {x - 2} \right) + \left( {x + 2} \right) = 0\\
\Rightarrow \left( {x + 2} \right)\left( {{x^2}\left( {x - 2} \right) + 1} \right) = 0\\
\Rightarrow \left( {x + 2} \right)\left( {{x^3} - 2{x^2} + 1} \right) = 0\\
\Rightarrow \left( {x + 2} \right)\left( {{x^3} - {x^2} - {x^2} + 1} \right) = 0\\
\Rightarrow \left( {x + 2} \right)\left( {x - 1} \right)\left( {{x^2} - \left( {x + 1} \right)} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x + 2 = 0\\
x - 1 = 0\\
{x^2} - x - 1 = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = - 2\left( {ktm} \right)\\
x = 1\left( {tm} \right)\\
{\left( {x - \frac{1}{2}} \right)^2} = \frac{5}{4}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 1\\
x = \frac{{1 \pm \sqrt 5 }}{2}\left( {tm} \right)
\end{array} \right.
\end{array}$
