$\iff 2x^3+2x^2+3x^2+3x-2x^3-5x^2=9$

=>3x=9

hay x=3

$\left(x^2-3x\right).\left(2x-1\right)-x.\left(2x^2-7x\right)=12$

$\iff 2x^3-x^2-6x^2+3x-x.\left(2x^2-7x\right)=12$

$\iff 2x^3-x^2-6x^2+3x-2x^3+7x^2=12$

$\iff -x^2-6x^2+3x+7x^2=12$

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

$\left(x+2\right).\left(x^2-4x+1\right)-\left(x^3-2x^2+2\right)=-18$

$\iff x^3-4x^2+x+2x^2-8x+2-\left(x^3-2x^2+2\right)=-18$

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

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

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

$\iff -7x=-18$

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

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=\(\dfrac{18}{7}\)