I also got Afrom 1) xy^2z^3=(xz)z^2 y^2
since xz>0, the square of any number >0
so, the equation will be ?0
from 2) xy^2z^3 = (xy)z^2 (yz)
since yz>0, z^2 >0, however, don't know x, y can either >0 or <0, so can't tell if this equation is >0 or <0
Any comments?
-- by 会员 SC2000 (2012/5/18 9:36:30)