ta có : \(x^2\left(x+1\right)=x\left(x+1\right)+x\left(x+1\right)=0\)
\(\Leftrightarrow x^2\left(x+1\right)=2x\left(x+1\right)=0\)
ta có : \(x^2\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x^2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=0\end{matrix}\right.\)
ta có : \(2x\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x+1=0\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=0\end{matrix}\right.\)
lấy dao nghiệm ta có \(x=-1;x=0\)
vậy \(x=-1;x=0\)
\(x^2.\left(x+1\right)=x.\left(x+1\right)+x.\left(x+1\right)=0\)
\(x^2.\left(x+1\right)=\left(x+1\right).\left(x+x\right)=0\)
\(x^2.\left(x+1\right)=\left(x+1\right).2x=0\)
\(\Rightarrow x^2.\left(x+1\right)-\left(x+1\right).2x=0\)
\(\left(x+1\right).\left(x^2-2x\right)=0\)
\(x.\left(x+1\right).\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=2\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=0\\x=-1\\x=2\end{matrix}\right.\)
Tham khảo nhé~
