a, Ta có : \(x^3-5x^2-2x+10=0\)
=> \(x^2\left(x-5\right)-2\left(x-5\right)=0\)
=> \(\left(x^2-2\right)\left(x-5\right)=0\)
=> \(\left[{}\begin{matrix}x^2-2=0\\x-5=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\pm\sqrt{2}\\x=5\end{matrix}\right.\)
Vậy ....
b, Ta có : \(x^3-6x^2+6x-1=0\)
=> \(x^3-x^2-5x^2+5x+x-1=0\)
=> \(\left(x-1\right)\left(x^2-5x+1\right)=0\)
=> \(\left[{}\begin{matrix}x=1\\x^2-\frac{2.x.5}{2}+\frac{25}{4}=\frac{21}{4}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=1\\x-\frac{5}{2}=\frac{\pm\sqrt{21}}{2}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=1\\x=\frac{5\pm\sqrt{21}}{2}\end{matrix}\right.\)
Vậy .....
