\(\frac{x-1}{5x+3}=\frac{3x-8}{x-1}\left(ĐKXĐ:x\ne\frac{-3}{5},x\ne1\right)\)
\(\Leftrightarrow\frac{\left(x-1\right)^2-\left(5x+3\right)\left(3x-8\right)}{\left(5x+3\right)\left(x-1\right)}\)
\(\Leftrightarrow x^2-2x+1-15x^2+31x+24=0\)
\(\Leftrightarrow-14x^2+29x+25=0\)
Pt nghiệm vô tỉ
Ta có: (x-1)(5x+3)=(3x-8)(x-1)
⇔(x-1)(5x+3)-(3x-8)(x-1)=0
⇔(x-1)(5x+3-3x+8)=0
hay (x-1)(2x+11)=0
⇔\(\left[{}\begin{matrix}x-1=0\\2x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\2x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{-11}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{1;\frac{-11}{2}\right\}\)
