Đáp án:
$17)
VT=xy(2x^3-3y^3)-x^2y^2(5x+4y)\\
=2x^4y-3xy^4-5x^3y^2-4x^2y^3\\
VP=2x^2y(x^2-xy+y^2)-3xy^2(x^2+2xy+y^2)\\
=2x^4y-2x^3y^2+2x^2y^3-3x^3y^2-6x^2y^3-3xy^4\\
=2x^4y+(-2x^3y^2-3x^3y^2)+(2x^2y^3-6x^2y^3)-3xy^4\\
=2x^4y-5x^3y^2-4x^2y^3-3xy^4=VT\Rightarrow đpcm\\
18)
VT=2y\left ( x^3+x^2y-\dfrac{1}{4}y^3 \right )-\dfrac{1}{2}x(2x^3+4xy^2-y^3)\\
=2x^3y+2x^2y^2-\dfrac{1}{2}y^4-x^4-2x^2y^2+\dfrac{1}{2}xy^3\\
=2x^3y+(2x^2y^2-2x^2y^2)-\dfrac{1}{2}y^4-x^4+\dfrac{1}{2}xy^3\\
=2x^3y-\dfrac{1}{2}y^4-x^4+\dfrac{1}{2}xy^3\\
VP= \dfrac{1}{2}y^3(x-y)-x^3(x-2y)\\
=\dfrac{1}{2}xy^3-\dfrac{1}{2}y^4-x^4+2x^3y=VT\Rightarrow đpcm\\$
