$(x-3)\left(\dfrac{80}{x}+10\right)=80+20$
$ĐK:x \neq 0$
$PT⇔ 10(x-3)+\dfrac{80(x-3)}{x}=100$
$⇔ 10x-30+\dfrac{80x-240}{x}=100$
$⇔ x(10x-30)+80x-240=100x$
$⇔ 10x^2-30x+80x-240=100x$
$⇔ 10x^2+50x-240=100x$
$⇔ 10x^2-50x-240=0$
$⇔ 10(x^2-5x-24)=0$
$⇔ x^2-5x-24=0$
$⇔ (x^2-8x)+(3x-24)=0$
$⇔ x(x-8)+3(x-8)=0$
$⇔ (x+3)(x-8)=0$
$⇔\left[ \begin{array}{l}x+3=0\Leftrightarrow x=-3\\x-8=0\Leftrightarrow x=8\end{array} \right.$
