ĐK: \(0< x< \sqrt{2}\)
Đặt \(x=a;\sqrt{2-x^2}=b\left(b>0\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{a}+\frac{1}{b}=2\\a^2+b^2=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=2ab\\a^2+b^2=2\end{matrix}\right.\)
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