Question

Answer 1

a) \(A=\dfrac{1}{2}x^2y^3\)

Hệ số: \(\dfrac{1}{2}\)

Biến: \(x^2y^3\)

Bậc: 5

\(B=\left(\dfrac{2}{3}xy\right)^2=\dfrac{4}{9}x^2y^2\)

Hệ số: \(\dfrac{4}{9}\)

Biến: \(x^2y^2\)

Bậc: 4

b) \(C=A.B=\dfrac{1}{2}x^2y^3.\dfrac{4}{9}x^2y^2=\dfrac{2}{9}x^4y^5\)

c) Thay x=1; y=-1 vào A, B

\(\left\{{}\begin{matrix}A=\dfrac{1}{2}.\left(1\right)^2.\left(-1\right)^3=-\dfrac{1}{2}\\B=\dfrac{4}{9}\left(1\right)^2.\left(-1\right)^2=\dfrac{4}{9}\end{matrix}\right.\)