Đáp án:
Giải thích các bước giải:
a) `x^3+3x^2+3x=0`
`⇔ x(x^2+3x+3)=0`
`⇔` \(\left[ \begin{array}{l}x=0\\x^2+3x+3=0 \ge 0 ∀x\end{array} \right.\)
Vậy `S={0}`
b) `x^3-3x^2+3x=0`
`⇔ x(x^2-3x+3)=0`
`⇔` \(\left[ \begin{array}{l}x=0\\x^2-3x+3=0 \ge 0 ∀x\end{array} \right.\)
Vậy `S={0}`
c) `x^3+6x^2+12x=0`
`⇔ x(x^2+6x+12)=0`
`⇔` \(\left[ \begin{array}{l}x=0\\x^2+6x+12=0 \ge 0 ∀x\end{array} \right.\)
Vậy `S={0}`
d) `x^3-6x^2+12x=0`
`⇔ x(x^2-6x+12)=0`
`⇔` \(\left[ \begin{array}{l}x=0\\x^2-6x+12=0 \ge 0 ∀x\end{array} \right.\)
Vậy `S={0}`
