`x^3-7x^2+12x-6=0`
`⇒x^3-x^2-6x^2+6x+6x-6=0`
`⇒(x^3-x^2)-(6x^2-6x)+(6x-6)=0`
`⇒x^2(x-1)-6x(x-1)+6(x-1)=0`
`⇒(x-1)(x^2-6x+6)=0`
`⇒`\(\left[ \begin{array}{l}x-1=0\\x^2-6x+6=0\end{array} \right.\)
`⇒`\(\left[ \begin{array}{l}x=1\\x^2-6x+9-9+6=0\end{array} \right.\)
`⇒`\(\left[ \begin{array}{l}x=1\\(x-3)^2-3=0\end{array} \right.\)
`⇒`\(\left[ \begin{array}{l}x=1\\(x-3-√3)(x-3+√3)=0\end{array} \right.\)
Ta có `(x-3-√3)(x-3+√3)=0`
`⇒`\(\left[ \begin{array}{l}x-3-√3=0\\x-3+√3\end{array} \right.\)
`⇒`\(\left[ \begin{array}{l}x=3+√3\\x=3-√3\end{array} \right.\)
Vậy \(\left[ \begin{array}{l}x=1\\x=3+√3\\x=3-√3\end{array} \right.\)
