Đáp án:
$\begin{array}{l}
Dkxd:4 - {x^2} \ge 0\\
\Rightarrow - 2 \le x \le 2\\
Đặt:x + \sqrt {4 - {x^2}} = t\\
\Rightarrow {x^2} + 2.x.\sqrt {4 - {x^2}} + 4 - {x^2} = {t^2}\\
\Rightarrow x.\sqrt {4 - {x^2}} = \dfrac{{{t^2} - 4}}{2}\\
\Rightarrow 3.x.\sqrt {4 - {x^2}} = \dfrac{{3{t^2} - 12}}{2}\\
Pt: \Rightarrow t = 2 + \dfrac{{3{t^2} - 12}}{2}\\
\Rightarrow 3{t^2} - 2t - 8 = 0\\
\Rightarrow 3{t^2} - 6t + 4t - 8 = 0\\
\Rightarrow \left( {t - 2} \right)\left( {3t + 4} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
t = 2\\
t = - \dfrac{4}{3}
\end{array} \right.\\
+ Khi:t = 2\\
\Rightarrow x.\sqrt {4 - {x^2}} = \dfrac{{{t^2} - 4}}{2} = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 0\\
x = 2\\
x = - 2
\end{array} \right.\left( {tmdk} \right)\\
+ Khi:t = - \dfrac{4}{3}\\
\Rightarrow x.\sqrt {4 - {x^2}} = \dfrac{{{t^2} - 4}}{2} = \dfrac{{ - 10}}{9}\\
\Rightarrow \left\{ \begin{array}{l}
- 2 \le x < 0\\
{x^2}\left( {4 - {x^2}} \right) = \dfrac{{100}}{{81}}
\end{array} \right.\\
\Rightarrow {x^4} - 4{x^2} + \dfrac{{100}}{{81}} = 0\\
\Rightarrow {x^2} = \dfrac{{22 + \sqrt {14} }}{9}\left( {ktm} \right)\\
Vậy\,x \in \left\{ {0; - 2;2} \right\}
\end{array}$
