\(x^2\le2\) , bình phương 2 vế"
\(10+2\sqrt{\left(2-x^2\right)\left(x^2+8\right)}=16\)
\(\Leftrightarrow\sqrt{-x^4-6x^2+16}=3\)
\(\Leftrightarrow x^4+6x^2-7=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=1\\x^2=-7< 0\left(l\right)\end{matrix}\right.\) \(\Rightarrow x=\pm1\)