ĐKXĐ: \(x\ge1;x\le-1\)
Đặt \(\sqrt{x^2-1}=y\left(y\ge0\right)\), phương trình trở thành:
\(x^2+3y=\sqrt{x^4-y^2}\)
\(\Leftrightarrow x^4+6x^2y+9y^2=x^4-y^2\)
\(\Leftrightarrow2y\left(5y+3x^2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=0\\5y+3x^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-1}=0\\5\sqrt{x^2-1}=-3x^2\left(\text{vô nghiệm}\right)\end{matrix}\right.\)
\(\Leftrightarrow x=\pm1\left(tm\right)\)