Đáp án:
$\begin{array}{l}
\sqrt { - 2{x^2} + 6} = x - 1\\
DKxd:\left\{ \begin{array}{l}
- 2{x^2} + 6 \ge 0\\
x - 1 \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{x^2} \le 3\\
x \ge 1
\end{array} \right.\\
\Rightarrow 1 \le x \le \sqrt 3 \\
pt \Rightarrow - 2{x^2} + 6 = {x^2} - 2x + 1\\
\Rightarrow 3{x^2} - 2x - 5 = 0\\
\Rightarrow \left( {3x - 5} \right)\left( {x + 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{5}{3}\left( {tm} \right)\\
x = - 1\left( {ktm} \right)
\end{array} \right.\\
Vậy\,x = \dfrac{5}{3}\\
\sqrt {x - 3} = 2\sqrt {2x + 1} \left( {dkxd:x \ge 3} \right)\\
\Rightarrow x - 1 = 4\left( {2x + 1} \right)\\
\Rightarrow x - 1 = 8x + 4\\
\Rightarrow 7x = - 5\\
\Rightarrow x = - \dfrac{5}{7}\left( {ktm} \right)\\
\Rightarrow pt\,vô\,nghiệm
\end{array}$
