Question

как решить пж 15 баллов

Answer 1

(3xy)^3*(1/3*xy^2)^2 = A; x = -3; y = 1/3.

1) 3xy = 3(-3)*1/3 = -3; (-3)^3 = -27

2) 1/3*xy^2 = 1/3*(-3)*(1/3)^2 = -1/9; (-1/9)^2 = 1/81

3) A = (-27)*1/81 = -27/81 = -1/3

Ответ: -1/3