Cách giải:
$x^6+x^5+x^4-17x^3+2x^2+4x+8=0$
$\to x^6-x^5+2x^5-2x^4+3x^4-3x^3-14x^3+14x^2-12x^2+12x-8x+8=0$
$\to x^5(x-1)+2x^4(x-1)+3x^3(x-1)-14x^2(x-1)-12x(x-1)-8(x-1)=0$
$\to (x-1)(x^5+2x^4+3x^3-14x^2-12x-8)=0$
$\to (x-1)(x^5-2x^4+4x^4-8x^3+11x^3-22x^2+8x^2-16x+4x-8)=0$
$\to (x-1)[x^4(x-2)+4x^3(x-2)+11x^2(x-2)+8x(x-2)+4(x-2)]=0$
$\to (x-1)(x-2)(x^4+4x^3+11x^2+8x+4)=0$
$x^4+4x^3+11x^2+8x+4$
$=x^4+4x^3+4x^2+7x^2+8x+4$
$=(x^2+x)^2+7(x^2+\dfrac{8}{7}x+\dfrac{16}{49})+\dfrac{12}{7}$
$=(x^2+x)^2+7(x+\dfrac{4}{7})^2+\dfrac{12}{7}>0$
$\to \left[ \begin{array}{l}x-1=0\\x-2=0\end{array} \right.$
$\to \left[ \begin{array}{l}x=1\\x=2\end{array} \right.$
Vậy $\left[ \begin{array}{l}x=1\\x=2\end{array} \right.$
