x=1 hoặc x = -11/2
$\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)<->\left(x-1\right)\left(5x+3\right)-\left(3x-8\right)\left(x-1\right)=0 <->\left(x-1\right)\left(5x+3-\left(3x-8\right)\right)=0 <->\left(x-1\right)\left(2x+11\right) <-> $\(\left[ \begin{array}{l}2x+11=0\\x-1=0\end{array} \right.\) <->\(\left[ \begin{array}{l}x=1\\x=-11/2\end{array} \right.\)