Đáp án:
$\begin{array}{l}
a)Dkxd:\left\{ \begin{array}{l}
5x + 10 \ge 0\\
8 - x \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x \ge - 2\\
x \le 8
\end{array} \right.\\
\Rightarrow - 2 \le x \le 8\\
\sqrt {5x + 10} = 8 - x\\
\Rightarrow 5x + 10 = {x^2} - 16x + 64\\
\Rightarrow {x^2} - 21x + 54 = 0\\
\Rightarrow {x^2} - 3x - 18x + 54 = 0\\
\Rightarrow \left( {x - 3} \right)\left( {x - 18} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 3\left( {tm} \right)\\
x = 18\left( {ktm} \right)
\end{array} \right.\\
\text{Vậy}\,x = 3\\
b)x - \sqrt {2x - 5} = 4\\
\Rightarrow \sqrt {2x - 5} = x - 4\left( {dk:x \ge 4} \right)\\
\Rightarrow 2x - 5 = {x^2} - 8x + 16\\
\Rightarrow {x^2} - 10x + 21 = 0\\
\Rightarrow {x^2} - 3x - 7x + 21 = 0\\
\Rightarrow \left( {x - 3} \right)\left( {x - 7} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 3\left( {ktm} \right)\\
x = 7\left( {tm} \right)
\end{array} \right.\\
\text{Vậy}\,x = 7\\
c)Dkxd:2 - x \ge 0 \Rightarrow x \le 2\\
\sqrt {{x^2} + 2x + 4} = 2 - x\\
\Rightarrow {x^2} + 2x + 4 = {x^2} - 4x + 4\\
\Rightarrow 6x = 0\\
\Rightarrow x = 0\left( {tmdk} \right)\\
\text{Vậy}\,x = 0\\
d)Dkxd:x \ge 2\\
\Rightarrow \sqrt {3{x^2} - 9x + 1} = x - 2\\
\Rightarrow 3{x^2} - 9x + 1 = {x^2} - 4x + 4\\
\Rightarrow 2{x^2} - 5x - 3 = 0\\
\Rightarrow 2{x^2} - 6x + x - 3 = 0\\
\Rightarrow \left( {x - 3} \right)\left( {2x + 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 3\left( {tm} \right)\\
x = - \frac{1}{2}\left( {ktm} \right)
\end{array} \right.\\
\text{Vậy}\,x = 3\\
e)Dkxd:x \ge 9\\
\sqrt {2x - 5} = x - 9\\
\Rightarrow 2x - 5 = {x^2} - 18x + 81\\
\Rightarrow {x^2} - 20x + 86 = 0\\
\Rightarrow {x^2} - 20x + 100 - 14 = 0\\
\Rightarrow {\left( {x - 10} \right)^2} = 14\\
\Rightarrow \left[ \begin{array}{l}
x = 10 + \sqrt {14} \left( {tm} \right)\\
x = 10 - \sqrt {14} \left( {ktm} \right)
\end{array} \right.\\
\text{Vậy}\,x = 10 + \sqrt {14}
\end{array}$
