a, (2x-3)²=9
⇔(2x-3-3)(2x-3+3)=0
⇔(2x-6).2x=0
⇔4x(x-3)=0
⇔\(\left[ \begin{array}{l}4x=0\\x-3=0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=0\\x=3\end{array} \right.\)
Vậy S={0;3}
b, (2x-5)²-(3x-1)²=0
⇔(2x-5-3x+1)(2x-5+3x-1)=0
⇔(-x-4)(5x-6)=0
⇔\(\left[ \begin{array}{l}-x-4=0\\5x-6=0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=-4\\x=6/5\end{array} \right.\)
Vậy S={-4;6/5}
