\(\left|x\right|\ge1\)
Đặt \(\sqrt[4]{x-\sqrt{x^2-1}}=a>0\Rightarrow\sqrt[4]{x+\sqrt{x^2-1}}=\frac{1}{a}\)
Phương trình trở thành:
\(a+\frac{1}{a^2}=2\Leftrightarrow a^3-2a^2+1=0\)
\(\Leftrightarrow\left(a-1\right)\left(a^2-a-1\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}a=1\\a=\frac{1+\sqrt{5}}{2}\\a=\frac{1-\sqrt{5}}{2}< 0\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt[4]{x-\sqrt{x^2-1}}=1\\\sqrt[4]{x-\sqrt{x^2-1}}=\frac{1+\sqrt{5}}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x-\sqrt{x^2-1}=1\\x-\sqrt{x^2-1}=\left(\frac{1+\sqrt{5}}{2}\right)^4=\frac{7+3\sqrt{5}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=\sqrt{x^2-1}\left(x\ge1\right)\\x-\frac{7+3\sqrt{5}}{2}=\sqrt{x^2-1}\left(x\ge\frac{7+3\sqrt{5}}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x+1=-1\\-\left(7+3\sqrt{5}\right)x+\frac{47+21\sqrt{5}}{2}=-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=7\end{matrix}\right.\)
