Đáp án:
$\begin{array}{l}
a)\sqrt {5x + 3} = 3x - 7\left( {dkxd:x \ge \frac{7}{3}} \right)\\
\Rightarrow 5x + 3 = {\left( {3x - 7} \right)^2}\\
\Rightarrow 5x + 3 = 9{x^2} - 42x + 49\\
\Rightarrow 9{x^2} - 42x + 46 = 0\\
\Rightarrow x = \frac{{7 + \sqrt 3 }}{3}\\
c)\sqrt {3{x^2} - 2x - 1} = 3x + 1\left( {dkxd:x \ge - \frac{1}{3}} \right)\\
\Rightarrow 3{x^2} - 2x - 1 = 9{x^2} + 6x + 1\\
\Rightarrow 6{x^2} + 8x + 2 = 0\\
\Rightarrow x = - \frac{1}{3}\\
e)x - \sqrt {2x - 5} = 4\\
\Rightarrow \sqrt {2x - 5} = x - 4\left( {dkxd:x \ge 4} \right)\\
\Rightarrow 2x - 5 = {x^2} - 8x + 16\\
\Rightarrow {x^2} - 10x + 21 = 0\\
\Rightarrow x = 7
\end{array}$
