$(2x−3)^2 =(2x-3)(x+1)$
$⇔$ $(2x−3)^2 -(2x-3)(x+1)=0$
$⇔$ $(2x−3)(2x-3-x-1)=0$
$⇔$ $(2x−3)(x-4)=0$
$⇔$ \(\left[ \begin{array}{l}2x-3=0\\x-4=0\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x=\frac{3}{2} \\x=4\end{array} \right.\)
Vậy $x∈ {\frac{3}{2};4}$
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Cho mk xin ctlhn nha! Thanks nhìu!