ĐKXĐ: \(\left[{}\begin{matrix}x=1\\x\le-4\\x\ge4\end{matrix}\right.\)
\(\sqrt{\left(x-1\right)\left(x+4\right)}-\sqrt{\left(x-1\right)\left(x-4\right)}-\left(x-1\right)=0\)
Nhận thấy \(x=1\) là 1 nghiệm
- Với \(x\ge4\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x+4}-\sqrt{x-4}-\sqrt{x-1}\right)=0\)
\(\Leftrightarrow\sqrt{x+4}=\sqrt{x-4}+\sqrt{x-1}\)
\(\Leftrightarrow x+4=2x-5+2\sqrt{x^2-5x+4}\)
\(\Leftrightarrow9-x=2\sqrt{x^2-5x+4}\) (\(x\le9\))
\(\Leftrightarrow x^2-18x+81=4\left(x^2-5x+4\right)\)
\(\Leftrightarrow3x^2-2x-65=0\Rightarrow\left[{}\begin{matrix}x=5\\x=-\frac{13}{3}\left(l\right)\end{matrix}\right.\)
- Với \(x\le-4\)
\(\Leftrightarrow\sqrt{1-x}\left(\sqrt{-x-4}-\sqrt{4-x}+\sqrt{1-x}\right)=0\)
\(\Leftrightarrow\sqrt{-x-4}+\sqrt{1-x}=\sqrt{4-x}\)
\(\Leftrightarrow-2x-3+2\sqrt{x^2+3x-4}=4-x\)
\(\Leftrightarrow2\sqrt{x^2+3x-4}=x-7\)
Do \(x\le-4\Rightarrow x-7< 0\Rightarrow ptvn\)
Vậy pt có 2 nghiệm \(x=\left\{1;5\right\}\)
