e, $\dfrac{3}{x+2}=1-\dfrac{2}{x-3}$ $(x\neq-2;3)$
$⇔\dfrac{3}{x+2}=\dfrac{x-5}{x-3}$
$⇒3(x-3)=(x+2)(x-5)$
$⇔3x-9=x²-3x-10$
$⇔x²-6x-1=0$
$⇔\left[ \begin{array}{l}x=3+\sqrt{10}(tm)\\x=3-\sqrt{10}(tm)\end{array} \right.$
f, $2x³-5x²+3x=0$
$⇔x.(2x²-5x+3)=0$
$⇔x.(x-1)(2x-3)=0$
$⇔\left[ \begin{array}{l}x=0\\x=1\\x=\dfrac{3}{2}\end{array} \right.$
g, $(2x+1)²-(x-1)²=0$
$⇔(x+2).3x=0$
$⇔\left[ \begin{array}{l}x=-2\\x=0\end{array} \right.$
