Question

Please answer and please don't cheat, it's urgent...​

Answer 1

Step-by-step explanation:

we know:

2 = K ^1/x

3 = K ^1/y

6 = 1/z

3×2 =6

x^a+x^b = x^a+b

substitute value of 3,2 and 6

K^1/x + K^ 1/y =K^1/z

The base are equal .

so,

1/x+1/y - 1/z

1/x+ 1/y+1/z =0

hope it helps you