$\dfrac{(x-3)(2x+2)}{x^2+2x+1}$ `= 0`
`⇔ (x−3)(2x+2)=0`
`⇔ 2x^2−4x−6=0`
`⇔ x^2−2x−3=0`
`⇔` $\left[ \begin{array}{l}a+b=-2\\ab=1(-3)=-3\end{array} \right.$
`⇔` $\left[ \begin{array}{l}a=-3\\b=1\end{array} \right.$
`⇔ (x^2−3x)+(x−3)`
`⇔ x(x−3)+x−3`
`⇔ (x−3)(x+1)`
$\left[ \begin{array}{l}x=3\\x=-1\end{array} \right.$
`x = 3`
Cho mình xin hay nhất nha bạn!!!!
@Drew Mclntyre
