$(x²-1)(x+2)(x-3)=(x-1)(x²-4)(x+5)$
$⇔(x-1)(x+1)(x+2)(x-3)-(x-1)(x-2)(x+2)(x+5)=0$
$⇔(x-1)(x+2).[(x+1)(x-3)-(x-2)(x+5)]=0$
$⇔(x-1)(x+2).(x²-3x+x-3-x²-5x+2x+10)=0$
$⇔(x-1)(x+2).(-5x+7)=0$
$⇔x-1=0$ hoặc $x+2=0$ hoặc $-5x+7=0$
$⇔x=1$ hoặc $x=-2$ hoặc $x=7/5$