Question

2x^{3} × 7x^{4} × (3x)^{2} =
(2p/q) × (-q)^{4} × 2/5p^{2} =
y^{5} x (x×y)^{3} x (1/x^{2} × y ) =

Answer 1

14 x^7 x dikali 9x^2=126 x^9
(2p/q) × q^4 × 2/5p^{2} =-4/5 p^3 q^3
y^5  x x^3y^3 x y/x^2=x^3y^8 x y/x^2=x y^8

Answer 2

14 x^7 x dikali 9x^2=126 x^9
(2p/q) × q^4 × 2/5p^{2} =-4/5 p^3 q^3
y^5  x x^3y^3 x y/x^2=x^3y^8 x y/x^2=x y^8