a, $x^3+1=x(x+1)$
$⇔(x+1)(x^2-x+1)-x(x+1)=0$
$⇔(x+1)(x^2-2x+1)=0$
$⇔(x+1)(x-1)^2=0$
\(⇔\left[ \begin{array}{l}x+1=0\\x-1=0\end{array} \right.⇔x=±1\)
Vậy ................
b, $x^3-7x+6=0$
$⇔(x^3-x)-(6x-6)=0$
$⇔x(x-1)(x+1)-6(x-1)=0$
$⇔(x-1)(x^2+x-6)=0$
$⇔(x-1)(x-2)(x+3)=0$
$⇔x-1=0$ hoặc $x-2=0$ hoặc $x+3=0$
$⇔x=1$ hoặc $x=2$ hoặc $x=-3$
Vậy ................
c, $x^6-x^2=0$
$⇔x^2(x^2-1)(x^2+1)=0$
$⇔x^2(x^2-1)=0($vì $x^2+1≥1>0∀x)$
\(⇔\left[ \begin{array}{l}x^2=0\\x^2-1=0\end{array} \right.⇔\)\(\left[ \begin{array}{l}x=0\\x=±1\end{array} \right.\)
Vậy ...............
d, $x^3-12=13x$
$⇔x^3-13x-12=0$
$⇔(x^3-x)-(12x+12)=0$
$⇔x(x-1)(x+1)-12(x+1)=0$
$⇔(x+1)(x^2-x-12)=0$
$⇔(x+1)(x+3)(x-4)=0$
$⇔x+1=0$ hoặc $x+3=0$ hoặc $x-4=0$
$⇔x=-1$ hoặc $x=-3$ hoặc $x=4$
Vậy .................
e, $-x^5+4x^4=-12x^3$
$⇔x^5-4x^4-12x^3=0$
$⇔x^3(x^2-4x-12)=0$
$⇔x^3(x+2)(x-6)=0$
$⇔x^3=0$ hoặc $x+2=0$ hoặc $x-6=0$
$⇔x=0$ hoặc $x=-2$ hoặc $x=6$
Vậy ..................
f, $x^3=4x$
$⇔x^3-4x=0$
$⇔x(x^2-4)=0$
\(⇔\left[ \begin{array}{l}x=0\\x^2-4=0\end{array} \right.\) \(⇔\left[ \begin{array}{l}x=0\\x=±2\end{array} \right.\)
Vậy ......................................
