Đáp án:
$\begin{array}{l}
x + 2\sqrt {2x - {x^2} + 3} = 1\\
Đkxd:2x - {x^2} + 3 \ge 0 \Rightarrow - 1 \le x \le 3\\
Pt \Rightarrow 2\sqrt {2x - {x^2} + 3} = 1 - x\\
\left( {đk:1 - x \ge 0 \Rightarrow x \le 1 \Rightarrow - 1 \le x \le 1} \right)\\
\Rightarrow 4\left( {2x - {x^2} + 3} \right) = {\left( {x - 1} \right)^2}\\
\Rightarrow 8x - 4{x^2} + 12 = {x^2} - 2x + 1\\
\Rightarrow 5{x^2} - 10x - 11 = 0\\
\Rightarrow \left[ \begin{array}{l}
x = \frac{{5 + 4\sqrt 5 }}{5}\left( {ktm} \right)\\
x = \frac{{5 - 4\sqrt 5 }}{5}\left( {tm} \right)
\end{array} \right.\\
Vậy\,x = \frac{{5 - 4\sqrt 5 }}{5}
\end{array}$
