Question

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Answer 1

Easy

2^a=3^b=6^(-c)=k

2^a=k

k^(1/a)=2

similarly

k^(1/b)=3

& k^(-1/c)=6

and since 2×3=6

[k^(1/a)]×[k^(1/b)]=k^(-1/c)

k^[(1/a)+(1/b)]=k^(-1/c)

since bases are same thats why powers will be equal

therefore, (1/a)+(1/b)= -1/c

(1/a)+(1/b)+(1/c)=0

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