a) \(A=\dfrac{1}{2}x^2y^3\) hệ số 1/2 biến :\(x^2y^3\) bậc : 5
\(B=\dfrac{4}{9}x^2y^2\) hệ số : 4/9 biến :\(x^2y^2\) bậc: 4
b)\(C=A.B\)
\(C=\dfrac{1}{2}x^2y^3\cdot\dfrac{4}{9}.x^2y^2=\dfrac{2}{9}x^4y^5\)
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
c) thay x = 1 , y = -1 vào A ta đc
\(A=\dfrac{1}{2}.1^2.\left(-1\right)^3=\dfrac{1}{2}.1.\left(-1\right)=-\dfrac{1}{2}\)
) thay x = 1 , y = -1 vào B ta đc
\(B=\dfrac{4}{9}.1^2.\left(-1\right)^2=\dfrac{4}{9}.1.1=\dfrac{4}{9}\)
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
\(\left\{{}\begin{matrix}A=\dfrac{1}{2}.\left(1\right)^2.\left(-1\right)^3=-\dfrac{1}{2}\\B=\dfrac{4}{9}\left(1\right)^2.\left(-1\right)^2=\dfrac{4}{9}\end{matrix}\right.\)