DK:\(-2\le x\le2\)
\(t=\sqrt{2+x}-2\sqrt{2-x}\Rightarrow t^2=10-3x-4\sqrt{4-x^2}\Rightarrow Pt\Leftrightarrow t^2=3t\Leftrightarrow\left[{}\begin{matrix}t=0\\t=3\end{matrix}\right.\)
\(t=0\Rightarrow\sqrt{2+x}=2\sqrt{2-x}\Leftrightarrow x=\dfrac{6}{5}\)
Thu lai thay x= 6/5 thoa man\(t=3\Rightarrow1-3x=4\sqrt{4-x^2}\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{1}{3}\\25x^2-6x-63=0\end{matrix}\right.\Leftrightarrow x=\dfrac{3-12\sqrt{11}}{25}}\)
ĐK: $ - 2 \leqslant x \leqslant 2$.
$3\sqrt {2 + x} - 6\sqrt {2 - x} + 4\sqrt {4 - {x^2}} = 10 - 3x$
Đặt $\left\{\begin{matrix} \sqrt{2 + x} = a\\ \sqrt {2 - x} = b \end{matrix}\right.$
PT $ \Leftrightarrow 3a - 6b + 4ab - {a^2} - 4{b^2} = 0$.
$\Leftrightarrow (a - 2b)(a - 2b - 3) = 0 \Leftrightarrow \left\{\begin{matrix} a=2b\\ a=2b+3 \end{matrix}\right.$ Nghiệm $x = \dfrac{6}{5}$.