Đặt \(x^3-4x^2+4x=0\)
=>\(x\left(x^2-4x+4\right)=0\)
=>\(x\left(x-2\right)^2=0\)
=>\(\left[{}\begin{matrix}x=0\\\left(x-2\right)^2=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Cho \(x^3-4x^2+4x=0\)
\(x\left(x^2-4x+4\right)=0\)
\(x\left(x^2-2x-2x+4\right)=0\)
\(x\left[\left(x^2-2x\right)-\left(2x-4\right)\right]=0\)
\(x\left[x\left(x-2\right)-2\left(x-2\right)\right]=0\)
\(x\left(x-2\right)\left(x-2\right)=0\)
\(x\left(x-2\right)^2=0\)
\(x=0\) hoặc \(\left(x-2\right)^2=0\)
*) \(\left(x-2\right)^2=0\)
\(x-2=0\)
\(x=2\)
Vậy nghiệm của đa thức đã cho là: \(x=0;x=2\)
