Đáp án:
$7)
a)
3x^2y^4-2x^3y^3-\dfrac{2}{9}xy^2\\
b)
\dfrac{2}{3}xy^3-\dfrac{4}{9}x^2y^2+x^2y^3\\
c)
\dfrac{-2}{3}x^3y^6+\dfrac{4}{9}x^4y^5\\
d)
\dfrac{-27}{2}xy^2+9x^2y$
Giải thích các bước giải:
$7)
a)
(-x^2y^3).(-3y+2x)+\left ( -\dfrac{2}{9}xy^2 \right )\\
=(-x^2y^3).(-3y)+(-x^2y^3).2x-\dfrac{2}{9}xy^2 \\
=3x^2y^4-2x^3y^3-\dfrac{2}{9}xy^2\\
b)
\left ( -\dfrac{2}{9}xy^2 \right ).(-3y+2x)-(-x^2y^3)\\
=\left ( -\dfrac{2}{9}xy^2 \right ).(-3y)+\left ( -\dfrac{2}{9}xy^2 \right ).2x+x^2y^3\\
=\dfrac{2}{9}.3xy^3-\dfrac{2}{9}.2x^2y^2+x^2y^3\\
=\dfrac{2}{3}xy^3-\dfrac{4}{9}x^2y^2+x^2y^3\\
c)
\left ( -\dfrac{2}{9}xy^2 \right ).(-3y+2x).(-x^2y^3)\\
=\left [ \dfrac{-2}{9}xy^2.(-3y)-\dfrac{2}{9}xy^2.2x \right ].(-x^2y^3)\\
=\left [ \dfrac{2}{3}xy^3-\dfrac{4}{9}x^2y^2\right ].(-x^2y^3)\\
=\dfrac{2}{3}xy^3.(-x^2y^3)-\dfrac{4}{9}x^2y^2.(-x^2y^3)\\
=\dfrac{-2}{3}x^3y^6+\dfrac{4}{9}x^4y^5\\
d)
\dfrac{-x^2y^3}{-\dfrac{2}{9}xy^2}.(-3y+2x)\\
=\dfrac{9}{2}xy.(-3y+2x)\\
=\dfrac{9}{2}xy.(-3y)+\dfrac{9}{2}xy.2x\\
=\dfrac{-27}{2}xy^2+9x^2y$
