Ta có : \(x^3-3x^2+4=0\)
=> \(x^3-2x^2-x^2+2x-2x+4=0\)
=> \(x^2\left(x-2\right)-x\left(x-2\right)-2\left(x-2\right)=0\)
=> \(\left(x^2-x-2\right)\left(x-2\right)=0\)
=> \(\left[{}\begin{matrix}x^2-x-2=0\\x-2=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x^2-x+\frac{1}{4}-\frac{9}{4}=0\\x-2=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left(x-\frac{1}{2}\right)^2=\frac{9}{4}\\x-2=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x-\frac{1}{2}=\pm\sqrt{\frac{9}{4}}\\x-2=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\sqrt{\frac{9}{4}}+\frac{1}{2}=2\\x=2\\x=-\sqrt{\frac{9}{4}}+\frac{1}{2}=-1\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
Vậy phương trình có 2 nghiệm là x = 2, x = -1 .
