Question

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

Answer 1

ĐKXĐ: x>=1 hoặc x<=-1

Ta có: \(\sqrt{x^2-1}-x^2+1=0\)

=>\(\sqrt{x^2-1}=x^2-1\)

=>\(\begin{cases}x^2-1\ge0\\ \left(x^2-1\right)^2=x^2-1\end{cases}\Rightarrow\begin{cases}x^2\ge1\\ \left(x^2-1\right)\left(x^2-1-1\right)=0\end{cases}\)

=>\(\begin{cases}x^2\ge1\\ \left(x^2-1\right)\left(x^2-2\right)=0\end{cases}\Rightarrow x^2\in\left\lbrace1;2\right\rbrace\)

=>\(x\in\left\lbrace1;-1;\sqrt2;-\sqrt2\right\rbrace\)