Lời giải:
Cách `1` :
`5(x-3)(x+2)=x^(2)-9`
`⇔(5x-15)(x+2)=x^(2)-9`
`⇔5x^(2)+10x-15x-30-x^(2)+9=0`
`⇔(5x^(2)-x^2)+(10x-15x)+(-30+9)=0`
`⇔4x^(2)-5x-21=0`
`⇔4x^(2)-12x+7x-21=0`
`⇔4x(x-3)+7(x-3)=0`
`⇔(x-3)(4x+7)=0`
$⇔\left[\begin{matrix}x-3=0\\ 4x+7=0\end{matrix}\right.$
$⇔\left[\begin{matrix}x=3\\ x=\dfrac{-7}{4}\end{matrix}\right.$
Vậy `x∈{3;-7/4}`
Cách `2` :
`5(x-3)(x+2)=x^(2)-9`
`⇔5(x-3)(x+2)-(x^(2)-3^2)=0`
`⇔5(x-3)(x+2)-(x-3)(x+3)=0`
`⇔(x-3)[5(x+2)-(x+3)=0`
`⇔(x-3)(5x+10-x-3)=0`
`⇔(x-3)(4x+7)=0`
$⇔\left[\begin{matrix}x-3=0\\ 4x+7=0\end{matrix}\right.$
$⇔\left[\begin{matrix}x=3\\ x=\dfrac{-7}{4}\end{matrix}\right.$
Vậy `x∈{3;-7/4}`
