高一数学问题，大神请进~O(∩_∩)O~

Also as 1/x +1/y = 1/z, z = 1/(1/x + 1/y)= xy/(x+y)

Then (ab)^z = (a * a^(x/y))^[xy/(x+y)] = a^[(1+ x/y)* xy / (x+y)] = a^[(x+y) / x * xy / (x+y)] = a^x

So a^x = (ab)^z.

As a^x = (ab)^z

Then a^x = (ab)^[xy/(x+y)]

So a = (ab)^[y / (x+y)], which leads to a^(x+y) = (ab)^y, then a^x *a^y = a^y *b^y,

Then we have a^x = b^y.

Recalling a^x = (ab)^z given in the condition, we have proven b^y = (ab)^z.