$(2x+7)²=9(x+2)²$
$⇔(2x+7)²-9(x+2)²=0$
$⇔(2x+7)²-(3x+6)²=0$
$⇔(2x+7-3x-6)(2x+7+3x+6)=0$
$⇔(-x-1)(5x+13)=0$
⇔\(\left[ \begin{array}{l}-x-1=0\\5x+13=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-1\\x=-13/5\end{array} \right.\)
vậy$x=-1$ $hoặc$ $x=13/5$
