Question

plz solve this question ....
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Answer 1

Step-by-step explanation:

2^x = 3^y = 6^-z = k

=> 2^x = k

=> 2 = k^1/x

=> 3^y = k

=> 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)

compare powers.

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

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

Answer 2

Answer:

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