$\dfrac{300}{x+5}+\dfrac53=\dfrac{345}{x}$
$ĐK: x \neq 0; x \neq -5$
$PT⇔ \dfrac{300.3x}{3x(x+5)}+\dfrac{5x(x+5)}{3x(x+5)}=\dfrac{345.3(x+5)}{3x(x+5)}$
$⇔ 900x+5x(x+5)=1035(x+5)$
$⇔ 900x+5x^2+25x=1035x+5175$
$⇔ 900x+5x^2+25x-1035x-5175=0$
$⇔ 5x^2-110x-5175=0$
$⇔ 5(x^2-22x-1035)=0$
$⇔ x^2-22x-1035=0$
$\Delta=b^2-4ac=484+4140=4624>0$
$⇒ \sqrt{\Delta}=68$
$\textrm{PT có 2 nghiệm phân biệt:}$
$x_1= \dfrac{-b+\sqrt{\Delta}}{2a}= \dfrac{22+68}{2}=45$
$x_2= \dfrac{-b-\sqrt{\Delta}}{2a}= \dfrac{22-68}{2}=-23$
