Question

\(x^2-3x-4=\sqrt{x-1}\left(x^2-4x-2\right)\)

Answer 1

ĐKXĐ: \(x\ge1\)

Đặt \(\left\{{}\begin{matrix}x^2-4x-2=a\\\sqrt{x-1}=b\end{matrix}\right.\)

\(\Rightarrow a+b^2-1=ab\)

\(\Leftrightarrow\left(b-1\right)\left(b+1\right)-a\left(b-1\right)=0\)

\(\Leftrightarrow\left(b-1\right)\left(b+1-a\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}b=1\\a-1=b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=1\\x^2-4x-3=\sqrt{x-1}\end{matrix}\right.\)