a, P= \(\left(\dfrac{-2}{3}x^3y^2\right)\left(\dfrac{1}{2}x^2y^5\right)\)
= \(\dfrac{-2}{3}x^3y^2.\dfrac{1}{2}x^2y^5\)
= \(\dfrac{-1}{3}x^5y^7\)
b, tại x= -1 y=1 ta co:
P= \(\dfrac{-1}{3}\left(-1\right)^5.1^7\) = 1/3
a)\(\left(\frac{-2}{3}x^3y^2\right)^2\left(\frac{1}{2}x^2y^5\right)=\frac{-4}{9}x^6y^4\frac{1}{2}x^2y^5=\frac{-2}{9}x^8y^9\)b) Thay x=-1 y=1 vào P ta được
P=\(\frac{-2}{9}\left(-1\right)^8\left(1\right)^9=\frac{-2}{9}\times1\times1=\frac{-2}{9}\)
a)Hệ số:\(\frac{-2}{9}\)
Phần biến :\(x^8y^9\)