求助，math几经123题，答案是不是错了啊，thx!!!

celinecoco1991

A  is for me, too,

tinalms0918

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)

我觉得这样正确的，关键在如何理解等式的表达。