ĐK:$x\epsilon\left[ {-2;2} \right]$(*)
pt$\Leftrightarrow -9x^{2}-8x+32+16\sqrt{2(x+2)(2-x)}=0$
$\Leftrightarrow 4(8-2x^{2})-8x-x^{2}+16\sqrt{8-2x^{2}}=0$(1)
Đặt $t=\sqrt{8-2x^{2}},t\geq0$
(1)tt $4t^{2}+16t-8x-x^{2}=0$
$\Leftrightarrow (2t+x+8)(2t-x)=0$
TH1:$t=\frac{-x-8}{2}\Leftrightarrow 2\sqrt{8-2x^{2}}=-x-8\Leftrightarrow \begin{cases}x\leq - 8\\ 9x^{2}-16x+32=0 \end{cases}$(VN)
TH2:$2t=x\Leftrightarrow \begin{cases}x\geq 0 \\ 9x^{2} =32\end{cases}\Leftrightarrow x=\frac{4\sqrt{2}}{3}$(t/m(*))
KL:....