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Given that a + b + c + d = 6 and 1/(a+b+c) + 1/(b+c+d) + 1/(a+b+d) + 1/(a+c+d) = 5/3, find the value of a/(b+c+d) + b/(a+c+d) + c/(a+b+d) + d/(a+b+c).

1. 4
2. 5
3. 6
4. 8

Correct answer: 6

Solution

Since a+b+c+d = 6, each three-variable sum equals 6 minus the excluded variable: b+c+d = 6-a, etc. So a/(b+c+d) = a/(6-a) = 6/(6-a) - 1. Summing: [a/(b+c+d) +... + d/(a+b+c)] = 6*[1/(6-a) + 1/(6-b) + 1/(6-c) + 1/(6-d)] - 4 = 6*(5/3) - 4 = 10 - 4 = 6.

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