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Question
If a and b are different positive primes such that
`(a+b)^-1(a^-1+b^-1)=a^xb^y,` find x + y + 2.
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Solution
`(a+b)^-1(a^-1+b^-1)=a^xb^y`
`rArr1/(a+b)(1/a+1/b)=a^xb^y`
`rArr1/(a+b)((a+b)/(ab))=a^xb^y`
`rArr(1/(ab))=a^xb^y`
`rArr(ab)^-1=a^xb6y`
`rArra^-1b^-1=a^xb^y`
⇒ x = -1 and y = -1
Therefore, the value of x + y + 2 is -1 - 1 + 2 = 0.
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