a. $x^3-2x^{2}y+xy-9x=x\left(x^2+2xy+y^2-9\right)=x{\left(x+y\right)}^2-3^2=x\left(x+y-3\right)\left(x+y+3\right)$

$=y\left(1-x^2-2xy-y^2\right)$

$=y\left[1-{\left(x+y\right)}^2\right]$

$=y\left(1-x-y\right)\left(1+y+x\right)$

(x2−3x).(2x−1)−x.(2x2−7x)=12⇔2x3−x2−6x2+3x−2x3+7x2=12⇔3x=12⇔x=4

$(x+2)(x^2-4x+1)-(x^3-2x^2+2)=-18$

$\iff x^3-4x^2+x+2x^2-8x+2-x^3+2x^2-2=18$

$\iff -7x=-18$

$\iff x=\frac{18}{7}$

Vậy $S=\left\{\frac{18}{7}\right\}$

(x+2).(x2−4x+1)−(x3−2x2+2)=−18⇔x3−4x2+x+2x2−8x+2−x3+2x2−2=−18⇔−7x=−18⇔x\(=\dfrac{18}{7}\)