ĐKXĐ: \(0< x< 4\)
Đặt \(\left\{{}\begin{matrix}\sqrt{2+\sqrt{x}}=a>0\\\sqrt{2-\sqrt{x}}=b\ge0\end{matrix}\right.\) \(\Rightarrow a^2+b^2=4\)
\(\Rightarrow\dfrac{a^2}{\sqrt{2}+a}+\dfrac{b^2}{\sqrt{2}-b}=\sqrt{2}\)
\(\Rightarrow a^2\sqrt{2}-a^2b+ab^2+b^2\sqrt{2}=2\sqrt{2}-2b+2a-ab\sqrt{2}\)
\(\Leftrightarrow\sqrt{2}\left(a^2+b^2\right)-ab\left(a-b\right)=2\sqrt{2}+2\left(a-b\right)-ab\sqrt{2}\)
\(\Leftrightarrow2\sqrt{2}+ab\sqrt{2}-ab\left(a-b\right)-2\left(a-b\right)=0\)
\(\Leftrightarrow\sqrt{2}\left(ab+2\right)-\left(a-b\right)\left(ab+2\right)=0\)
\(\Leftrightarrow\left(\sqrt{2}-a+b\right)\left(ab+2\right)=0\)
\(\Leftrightarrow\sqrt{2}-a+b=0\) (do \(ab\ge0\Rightarrow ab+2>0\))
\(\Leftrightarrow\sqrt{2+\sqrt{x}}-\sqrt{2-\sqrt{x}}=\sqrt{2}\)
Hiển nhiên \(2+\sqrt{x}\ge2-\sqrt{x}\) nên:
\(\Leftrightarrow2+\sqrt{x}+2-\sqrt{x}-2\sqrt{4-x}=2\)
\(\Leftrightarrow\sqrt{4-x}=1\)
\(\Rightarrow x=3\)
