$a)x^2-4+(x+2)(x-3)=0$
$->[(x^2-(2)^2]+(x+2)(x-3)=0$
$->(x-2)(x+2)+(x+2)(x-3)=0$
$->(x+2)[x-2+(x-3)]$
$->(x+2)(2x-5)$
Nếu $x+2=0–> x=-2$
hoặc $ 2x-5=0–> x=(5)/2$
$b)(3x+1)^2-16=0$
$->(3x+1)^2-4^2=0$( do đấy là $HĐT$)
$-> (3x+1-4)(3x+1+4)=0$
$->(3x-3)(3x+5)=0$
$->3(x-1)(3x+5)=0$
Nếu $x-1=0–> x=1$
hoặc $ 3x+5=0–> x=(-5)/3$
$#Học tốt$