Đáp án:
`1)S={-2/3;1/2}`
`2)S={-3/2;6}`
Giải thích các bước giải:
`1)9x²-4-(3x+2)(x-1)=0`
`⇔(3x+2)(3x-2)-(3x+2)(x-1)=0`
`⇔(3x+2)[(3x-2)-(x-1)]=0`
`⇔(3x+2)(3x-2-x+1)=0`
`⇔(3x+2)(2x-1)=0`
`⇔`\(\left[ \begin{array}{l}3x+2=0\\2x-1=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}3x=-2\\2x=1\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=-\dfrac{2}{3}\\x=\dfrac{1}{2}\end{array} \right.\)
Vậy `S={-2/3;1/2}`
`2)(x+3)(2x+3)=4x²-9`
`⇔(x+3)(2x+3)=(2x+3)(2x-3)`
`⇔(x+3)(2x+3)-(2x+3)(2x-3)=0`
`⇔(2x+3)[(x+3)-(2x-3)]=0`
`⇔(2x+3)(x+3-2x+3)=0`
`⇔(2x+3)(-x+6)=0`
`⇔`\(\left[ \begin{array}{l}2x+3=0\\-x+6=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}2x=-3\\-x=-6\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=-\dfrac{3}{2}\\x=6\end{array} \right.\)
Vậy `S={-3/2;6}`