ĐK:\(-2\le x\le2\)
\(\Leftrightarrow\sqrt{2-x}+\sqrt{2+x}+\sqrt{\left(2+x\right)\left(2-x\right)}=2\)
Đặt \(a=\sqrt{2-x};b=\sqrt{2+x}\left(a,b\ge0\right)\)
\(\Rightarrow\left\{{}\begin{matrix}a^2+b^2=4\\a+b+ab=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(a+b\right)^2-2ab=4\\a+b+ab=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(2-ab\right)^2-2ab=4\\a+b=2-ab\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a^2b^2+4-6ab=4\\a+b=2-ab\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}ab=6\\a+b=-4\end{matrix}\right.\\\left\{{}\begin{matrix}ab=0\\a+b=2\end{matrix}\right.\end{matrix}\right.\)
Vậy a;b là ng0 của pt: \(\left[{}\begin{matrix}X^2+4X+6=0\left(vl\right)\\X^2-2X=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}X=0\\X=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=0\\b=2\end{matrix}\right.\\\left\{{}\begin{matrix}a=2\\b=0\end{matrix}\right.\end{matrix}\right.\)
-TH 1:\(\Rightarrow\left\{{}\begin{matrix}\sqrt{2-x}=0\\2+x=4\end{matrix}\right.\)\(\Rightarrow x=2\)
-TH 2:\(\Rightarrow\left\{{}\begin{matrix}2-x=4\\2+x=0\end{matrix}\right.\)(vô lí)
Vậy pt có tập ng0 S={2}