ĐKXĐ: \(x\ne1\)
\(\dfrac{x^2-4x}{x-1}\left(x+\dfrac{x-4}{x-1}\right)=5\Leftrightarrow\dfrac{x^2-4x}{x-1}\left(\dfrac{x^2-4}{x-1}\right)=5\)
\(\Leftrightarrow\dfrac{x^2-4x}{x-1}\left(\dfrac{x^2-4x}{x-1}+4\right)=5\)
Đặt \(\dfrac{x^2-4x}{x-1}=a\) phương trình trở thành:
\(a\left(a+4\right)-5=0\Leftrightarrow a^2+4a-5=0\)
\(\Rightarrow\left[{}\begin{matrix}a=1\\a=-5\end{matrix}\right.\)
- Với \(a=1\Rightarrow\dfrac{x^2-4x}{x-1}=1\Leftrightarrow x^2-4x=x-1\)
\(\Leftrightarrow x^2-5x+1=0\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5+\sqrt{21}}{2}\\x=\dfrac{5-\sqrt{21}}{2}\end{matrix}\right.\)
- Với \(a=-5\Rightarrow\dfrac{x^2-4x}{x-1}=-5\Leftrightarrow x^2-4x=-5x+5\)
\(\Leftrightarrow x^2+x-5=0\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1-\sqrt{21}}{2}\\x=\dfrac{-1+\sqrt{21}}{2}\end{matrix}\right.\)
Vậy pt đã cho có 4 nghiệm ....
