Question

giải phương trình \(\sqrt{x^2-1}-x^2+1=0\)

Answer 1

ĐK:\(\left[{}\begin{matrix}x\ge1\\x\le-1\end{matrix}\right.\)

\(\sqrt{x^2-1}-x^2+1=0\Leftrightarrow\sqrt{x^2-1}-\left(x^2-1\right)=0\Leftrightarrow\sqrt{x^2-1}\left(1-\sqrt{x^2-1}\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}\sqrt{x^2-1}=0\\1-\sqrt{x^2-1}=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x^2=1\\x^2=2\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=\pm1\left(tm\right)\\x=\pm\sqrt{2}\left(tm\right)\end{matrix}\right.\)

Vậy S={\(\pm1;\pm\sqrt{2}\)}