Giải thích các bước giải:
$a.(3x+2)(x-4)=0$
$\to 3x+2=0\to x=-\dfrac 23$
Hoặc $x-4=0\to x=4$
b.$\dfrac x2+\dfrac{4x-2}{4}=\dfrac{2x+2}{3}$
$\to \dfrac{2x}4+\dfrac{4x-2}{4}=\dfrac{2x+2}{3}$
$\to\dfrac{2x+4x-2}{4}=\dfrac{2x+2}{3}$
$\to\dfrac{6x-2}{4}=\dfrac{2x+2}{3}$
$\to\dfrac{3x-1}{4}=\dfrac{x+1}{3}$
$\to 3(3x-1)=4(x+1)$
$\to 9x-3=4x+4$
$\to 5x=7$
$\to x=\dfrac 75$