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Simplify (x/7)^2*[(7/x)^-3]^5=(x/7)^2p-1

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Solution: we have,

$\ (x/7)^2 imes [(7/x)^-3]^5=(x/7)^2p-1\ \Rightarrow x^17/7^17=(x/7)^2p-1\ \ 2p-1=17\ \Rightarrow 2p=18\ \Rightarrow p=9$

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Deependra Verma

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