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