Giải:
a)
$(x-1).(5x+3)=(3x-8).(x-1)$
$<=>(x-1).(5x+3)-(3x-8).(x-1)=0$
$<=>(x-1).(5x+3-(3x-8))=0$
$<=>(x-1).(2x+11)=0$
<=>\(\left[ \begin{array}{l}x-1=0\\2x+11=0\end{array} \right.\)
<=>\(\left[ \begin{array}{l}x=1\\x=\frac{-11}{2}\\\end{array} \right.\)
Vậy \(\left[ \begin{array}{l}x=1\\x=\frac{-11}{2}\\\end{array} \right.\)
b.
$(2-3x).(x+11)=(3x-2).(2-5x)$
$<=>-(3x-2).(x+11)=(3x-2).(2-5x)$
$<=>-(3x-2).(x+11)-(3x-2).(2-5x)=0$
$<=>(3x-2).(-x-11-(2-5x))=0$
$<=>(3x-2).(4x-13)=0$
<=>\(\left[ \begin{array}{l}3x-2=0\\4x-13=0\end{array} \right.\)
<=>\(\left[ \begin{array}{l}x=\frac{2}{3}\\x=\frac{13}{4}\\\end{array} \right.\)
Vậy \(\left[ \begin{array}{l}x=\frac{2}{3}\\x=\frac{13}{4}\\\end{array} \right.\)
c.
$(2x-1)^2-9.(2x-1)^2=0$
$<=>-8.(2x-1)^2=0$
$<=>(2x-1)^2=0$
$<=>2x-1=0$
$<=>2x=1$
$<=>x=\frac{1}{2}$
Vậy $x=\frac{1}{2}$
