a)$(x-1)(x+3)=x^2-4$
$⇔x^2+3x−x−3=x^2−4$
$⇔x^2+2x−3=x^2−4$
$⇔2x−3=−4$
$⇔2x=−4+3$
$⇔2x=−1$
$⇔x=-\dfrac{1}{2}$
b)$(x-2)(x-5)=(x-3)(x-4)$
$⇔x^2−5x−2x+10=x^2−4x−3x+12$
$⇔x^2−7x+10=x^2−7x+12$
$⇔10=12$(vô nghiệm)
$x ∈ ∅$
c)$(6x+2) (x-2) = 2x(3x-5)$
$⇔6x^2−12x+2x−4=6x^2−10x$
$⇔6x^2−10x−4=6x^2−10x$
$⇔−4=0 $(vô nghiệm)
$⇔x ∈ ∅$
d)$(x-1)(x+3)-(x+2)(x-3)=0$
$⇔x^2+3x−x−3−x^2+3x−2x+6=0$
$⇔3x+3=0$
$⇔x=−1$