Question

please solve this....
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Answer 1

Answer:

yes it is true

Step-by-step explanation:

let every value be k then

                          2= k^1/x

                          3= k^1/y

                          6= k^-1/z

2*3=6

k^1/x*k^1/y= k^-1/z

k^(1/x+1/y)=k^-1/z

1/x+1/y= -1/z

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