`c) (x^2-1)(x+3)=0 `
`<=>(x+1)(x-1)(x+3)=0`
`<=>x+1=0` hoặc `x-1=0` hoặc `x+3=0`
`<=>x=-1` hoặc `x=1` hoặc `x=-3`
Vậy `S={-1;1;3}`
`d) (x^2+1)(x^2+4x+4)=0`
`<=>(x^2+1)(x+2)^2=0`
`<=>`\(\left[ \begin{array}{l}x^2+1=0\\(x+2)^2=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x^2=-1( vô lí ) \\x+2=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x^2=-1( vô lí )\\x=-2\end{array} \right.\)
Vậy `S={-2}`
`e) x^2-x-6=0`
`<=>x^2-3x+2x-6=0`
`<=>(x^2-3x)+(2x-6)=0`
`<=>x(x-3)+2(x-3)=0`
`<=>(x-3)(x+2)`
`<=>`\(\left[ \begin{array}{l}x-3=0\\x+2=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=3\\x=-2\end{array} \right.\)
Vậy `S={3;-2}`
