a,$x^3+27+\left(x+3\right)\left(x-9\right)=0$

$x^3+27+x^2-6x-27=0$

$x^3+x^2-6x=0$

$x^3-2x^2+3x^2-6x=0$

$\left(x^2+3\right)\left(x-2\right)=0$

Vì $x^2+3>0$ nên $x-2=0=>x=2$

Vậy...

b,$x^3-11x^2+30x=0$

$x^3-6x^2-5x^2+30x=0$

$\left(x-6\right)\left(x^2-5\right)=0$

=>$x-6=0$ hoặc $x^2-5=0$

=>$x=6$hoặc $x^2=5$

=>$x=6$ hoặc $x=-\sqrt{5}$ hoặc $x=\sqrt{5}$

Giải các phương trình sau:

a,$\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6$

b,$4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10$

a)$\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6$

$\Leftrightarrow x^2-4x+4-x^2+9-6=0$

$\Leftrightarrow-4x+7=0$$\Leftrightarrow x=\dfrac{7}{4}$

Vậy $S=\left\{\dfrac{7}{4}\right\}$

b)$4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10$

$\Leftrightarrow4\left(x^2-6x+9\right)-4x^2+1-10=0$

$\Leftrightarrow4x^2-24x+36-4x^2+1-10=0$

$\Leftrightarrow-24x+27=0\Leftrightarrow x=\dfrac{9}{8}$

Vậy $S=\left\{\dfrac{9}{8}\right\}$