a) `3x+2(x-7)=-3`
`⇔3x+2x-14=-3`
`⇔5x=11`
`⇔x=11/5`
b) `(2x-3)^2-(x-1)^2=0`
`⇔(2x-3-x+1)(2x-3+x-1)=0`
`⇔(x-2)(3x-4)=0`
\(⇔\left[ \begin{array}{l}x-2=0\\3x-4=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=2\\x=\dfrac{4}{3}\end{array} \right.\)
Vậy `x=2; 4/3`