Lời giải:
`a)`
`2(x-3)=x^(2)-9`
`⇔2(x-3)=(x-3)(x+3)`
`⇔2(x-3)-(x-3)(x+3)=0`
`⇔(x-3)(2-x-3)=0`
`⇔(x-3)(-x-1)=0`
$⇔\left[\begin{matrix}x-3=0\\ -x-1=0\end{matrix}\right.$
$⇔\left[\begin{matrix}x=3\\ x=-1\end{matrix}\right.$
Vậy `x∈{3;-1}`
`b)`
`(2-3x)^(2)-4x^2=0`
`⇔(2-3x)^(2)-(2x)^2=0`
`⇔(2-3x-2x)(2-3x+2x)=0`
`⇔(2-5x)(2-x)=0`
$⇔\left[\begin{matrix}2-5x=0\\ 2-x=0\end{matrix}\right.$
$⇔\left[\begin{matrix}5x=2\\ x=2\end{matrix}\right.$
$⇔\left[\begin{matrix}x=\dfrac{2}{5}\\ x=2\end{matrix}\right.$
Vậy `x={2/5;2}`
