Ta có: \(x^3+6x^2+9x=0\)
\(\Leftrightarrow x\left(x+3\right)^2=0\)
hay \(x\in\left\{0;-3\right\}\)
x3+6x2+9x=0
⇒x(x2+6x+9)=0
⇒x(x+3)2=0
⇒\(\left[{}\begin{matrix}x=0\\\left(x+3\right)^2=0\end{matrix}\right.\)
⇒\(\left[{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.\)
⇒\(\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
\(pt< =>x\left(x^2+6x+9\right)=0< =>x\left(x+3\right)^2=0\)
\(=>\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
x3 + 6x2 + 9x =0
x3 + 3x2 + 3x2 + 9x=0
x2(x+3) + 3x(x+3) =0
(x+3)(x2+3x)=0
x(x+3)(x+3)=0
x(x+3)2 =0
+) x =0
+) (x+3)2 =0
\(\Rightarrow\)x+3=0
\(\Rightarrow\)x=-3
Vậy x =-3 hoặc x =0
