ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\sqrt{2+\sqrt{x}}=a>0\\\sqrt{2-\sqrt{x}}=b\ge0\end{matrix}\right.\) \(\Rightarrow a^2+b^2=4\)
\(\dfrac{a^2}{\sqrt{2}+a}+\dfrac{b^2}{\sqrt{2}-b}=\sqrt{2}\)
\(\Rightarrow a^2\left(\sqrt{2}-b\right)+b^2\left(\sqrt{2}+a\right)=\sqrt{2}\left(\sqrt{2}+a\right)\left(\sqrt{2}-b\right)\)
\(\Leftrightarrow\sqrt{2}\left(a^2+b^2+ab-2\right)=\left(a-b\right)\left(ab+2\right)\)
\(\Leftrightarrow\sqrt{2}\left(ab+2\right)=\left(a-b\right)\left(ab+2\right)\)
\(\Leftrightarrow a-b=\sqrt{2}\)
\(\Leftrightarrow\sqrt{2+\sqrt{x}}-\sqrt{2-\sqrt{x}}=\sqrt{2}\)
Do \(2+\sqrt{x}>2-\sqrt{x}\Rightarrow\) vế trái luôn dương, bình phương:
\(\Rightarrow4-2\sqrt{4-x}=2\)
\(\Leftrightarrow x=3\)
