Đáp án:
$\begin{array}{l}
1)\sqrt {x + 1} = \sqrt 5 - 3\\
Do:\left\{ \begin{array}{l}
\sqrt {x + 1} \ge 0\\
\sqrt 5 - 3 < 0
\end{array} \right.\\
\text{Vậy pt vô nghiệm}\\
2)Dkxd:x \ge \dfrac{2}{{25}}\\
\sqrt {25x - 2} = 10\\
\Leftrightarrow 25x - 2 = 100\\
\Leftrightarrow 25x = 102\\
\Leftrightarrow x = \dfrac{{102}}{{25}}\left( {tm} \right)\\
\text{Vậy}\,x = \dfrac{{102}}{{25}}\\
3)Dkxd:x \le 0\\
\sqrt { - 6x} = \sqrt 5 \\
\Leftrightarrow - 6x = 5\\
\Leftrightarrow x = \dfrac{{ - 5}}{6}\left( {tmdk} \right)\\
\text{Vậy}\,x = \dfrac{{ - 5}}{6}\\
4)\sqrt {{{\left( {2 - x} \right)}^2}} = 3\\
\Leftrightarrow \left| {2 - x} \right| = 3\\
\Leftrightarrow \left[ \begin{array}{l}
2 - x = 3 \Leftrightarrow x = - 1\\
2 - x = - 3 \Leftrightarrow x = 5
\end{array} \right.\\
\text{Vậy}\,x = - 1;x = 5\\
5)Dkxd:x \ge 1\\
\sqrt {{x^2} + 2x + 2} = x - 1\\
\Leftrightarrow {x^2} + 2x + 2 = {x^2} - 2x + 1\\
\Leftrightarrow 4x = - 1\\
\Leftrightarrow x = - \dfrac{1}{4}\left( {ktm} \right)\\
\text{Vậy pt vô nghiệm}
\end{array}$
