Đáp án + Giải thích các bước giải:
a) $4(x+3)-6x=2$
$⇔4x+12-6x-2=0$
$⇔10-2x=0$
$⇔x=5$
b) $4(x-2)+9x(x-2)=0$
$⇔(x-2)(4+9x)=0$
$⇔\left[\begin{matrix} x=2\\ x=-\dfrac{4}{9}\end{matrix}\right.$
c) ĐKXĐ: $x\neq±6$
$\dfrac{7}{x-6}+\dfrac{2x}{x+6}=\dfrac{2x(x-4)}{x^2-36}$
$⇔\dfrac{7(x+6)}{(x-6)(x+6)}+\dfrac{2x(x-6)}{(x-6)(x+6)}=\dfrac{2x(x-4)}{(x-6)(x+6)}$
$⇔\dfrac{7x+42+2x^2-12x-2x^2+8x}{(x-6)(x+6)}=0$
$⇔3x+42=0$
$⇔x=-14$
