Bài 1:
1. $3(x-1)-4=2(x+1)-7$
⇔ $3x-3-4=2x+2-7$
⇔ $3x-2x=2-7+3+4$
⇔ $x=2$
Vậy $S=${$2$}
2. $(x+2)^2-3(x-1)=(x+1)(x-2)$
⇔ $x^2+4x+4-3x+3=x^2-2x+x-2$
⇔ $x^2+x+7=x^2-x-2$
⇔ $x^2-x^2+x+x=-2-7$
⇔ $2x=-9$
⇔ $x=\frac{-9}{2}$
Vậy $S=${$\frac{-9}{2}$}
Bài 2:
1. $(3x-2)(4x+5)=0$
⇔ \(\left[ \begin{array}{l}3x-2=0\\4x+5=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=\frac{2}{3}\\x=\frac{-5}{4}\end{array} \right.\)
Vậy $S=${$\frac{2}{3};\frac{-5}{4}$}
2. $(2x-3)^2=(2x-3)(x+1)$
⇔ $4x^2-12x+9=2x^2+2x-3x-3$
⇔ $4x^2-2x^2-12x-2x+3x+3+9=0$
⇔ $2x^2-11x+12=0$
⇔ $(x-4)(2x-3)=0$
⇔ \(\left[ \begin{array}{l}x-4=0\\2x-3=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=4\\x=\frac{3}{2}\end{array} \right.\)
Vậy $S=${$\frac{3}{2};4$}
