Đặt \(a=\sqrt{2+\sqrt{x}};b=\sqrt{2-\sqrt{x}}\left(a,b\ge0\right)\Rightarrow a^2+b^2=4\)
Khi đó, ta thu được pt sau: \(\frac{a^2}{\sqrt{2}+a}+\frac{b^2}{\sqrt{2}-b}=\sqrt{2}\)
\(\Leftrightarrow\frac{\sqrt{2}\left(a^2+b^2\right)-ab\left(a-b\right)}{\left(\sqrt{2}+a\right)\left(\sqrt{2}-b\right)}=\sqrt{2}\)
\(\Rightarrow4\sqrt{2}-ab\left(a-b\right)=\sqrt{2}\left(2+a\sqrt{2}-b\sqrt{2}-ab\right)\) (Vì a2+b2=4)
\(\Leftrightarrow2\sqrt{2}-ab\left(a-b\right)-2\left(a-b\right)+ab\sqrt{2}=0\)
\(\Leftrightarrow\sqrt{2}\left(ab+2\right)-\left(a-b\right)\left(ab+2\right)=0\)
\(\Leftrightarrow\left(ab+2\right)\left(\sqrt{2}-a+b\right)=0\Leftrightarrow\orbr{\begin{cases}ab+2=0\\b+\sqrt{2}=a\end{cases}}\)(loại \(ab+2=0\) vì \(ab\ge0\))
\(\Leftrightarrow b+\sqrt{2}=a\Rightarrow\sqrt{2-\sqrt{x}}+\sqrt{2}=\sqrt{2+\sqrt{x}}\)
\(\Leftrightarrow2-\sqrt{x}+2+2\sqrt{4-2\sqrt{x}}=2+\sqrt{x}\)
\(\Leftrightarrow2-2\sqrt{x}+2\sqrt{4-2\sqrt{x}}=0\Leftrightarrow\sqrt{4-2\sqrt{x}}=\sqrt{x}-1\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{x}-1\ge0\\4-2\sqrt{x}=x-2\sqrt{x}+1\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge1\\x=3\left(tm\right)\end{cases}}\)
Vậy pt cho có nghiệm duy nhất x=3.
