a, -3x^2y.2/3xy^2 = -2x^3y^4
b,-2x^2y.2/3x^2y^2 = =4/3x^4y^3
bạn chỉ cần nhân từng phần với nhau là được rồi
\(a,-3x^2y\cdot\frac{2}{3}xy^3\)
\(=\left(-3\cdot\frac{2}{3}\right)\left(x^2yxy^3\right)\)
\(=-2\left(x^2x\right)\left(yy^3\right)\)
\(=-2x^3y^4\)
Hok tốt {} {}
=^.^=
\(b,-2x^2y\cdot\frac{2}{3}x^2y^2\)
\(=\left(-2\cdot\frac{2}{3}\right)\left(x^2ỹ^{ }x^2y^2\right)\)
\(=-\frac{4}{3}\left(x^2x^2\right)\left(yy^2\right)\)
\(=-\frac{4}{3}x^4y^3\)
`````Hok Tốt ````
a) \(-3x^2y.\frac{2}{3}xy^3=\left(-3.\frac{2}{3}\right).\left(x^2y.xy^3\right)=-2.\left(x^2x.yy^3\right)=-2x^3y^4\)
b) \(-2x^2y.\frac{2}{3}.x^2y^2=\left(-2.\frac{2}{3}\right).\left(x^2y.x^2y^2\right)=-\frac{4}{3}.\left(x^2x^2.yy^2\right)=-\frac{4}{3}x^4y^3\)
\(-3x^2y.\frac{2}{3}xy^3\)
\(=\left(-3.\frac{2}{3}\right).\left(x^2y.xy^3\right)\)
\(=-2x^3y^4\)
\(-2x^2y.\frac{2}{3}x^2y^2\)
\(=\left(-2.\frac{2}{3}\right).\left(x^2y.x^2y^2\right)\)
\(=-\frac{4}{3}x^4y^3\)
\(a,-2x^3y^4\)
\(b,-\frac{4}{3}x^4y^3\)
hok tốt