a) `x(x+3)-2(x+3)=0`
⇔`(x+3)(x-2)=0`
⇔\(\left[ \begin{array}{l}x+3=0\\x-2=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-3\\x=2\end{array} \right.\)
Vậy `S={-3,2}`
b) Sửa đề
`(x-2)(x^2+2x+7)+2(x^2-4)-5(x-2)=0`
⇔`(x-2)(x^2+2x+7+2x+4-5)=0`
⇔`(x-2)(x^2+4x-6)=0`
Mà `x^2+4x-6≥2∀x`
⇒`x-2=0`
⇔`x=2`
Vậy `S={2}`