Đáp án + Giải thích các bước giải:
`a//x(x-3)+7(x-3)=0`
`⇔(x-3)(x+7)=0`
`⇔` \(\left[ \begin{array}{l}x-3=0\\x+7=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=3\\x=-7\end{array} \right.\)
Vậy `S={3;-7}`
`b//x^{2}(x+2)-9(x+2)=0`
`⇔(x+2)(x^{2}-9)=0`
`⇔(x+2)(x+3)(x-3)=0`
`⇔` \(\left[ \begin{array}{l}x+2=0\\x+3=0\\x-3=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-2\\x=-3\\x=3\end{array} \right.\)
Vậy `S={-2;±3}`
`c//x^{2}(x+3)=4(x+3)`
`⇔x^{2}(x+3)-4(x+3)=0`
`⇔(x+3)(x^{2}-4)=0`
`⇔(x+3)(x+2)(x-2)=0`
`⇔` \(\left[ \begin{array}{l}x+3=0\\x+2=0\\x-2=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-3\\x=-2\\x=2\end{array} \right.\)
Vậy `S={-3;±2}`
