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Can anyone solve this...? Steps needed.. If |a-b| =1 ; |b-c| =1 ; |c-a| = 2 and abc= 60Find a/bc + b/ca + c/ab - 1/a - 1/b - 1/c |
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Answer» - b| = 1|b - c| = 1|c - a| = 1abc = 60To finda/bc + b/ca + c/ab - 1/a - 1/b - 1/c SolutionTo remove modulus, square both sides|a - b|² = 1²→ a² + b² - 2ab = 1|b - c|² = 1²→ b² + c² - 2BC = 1|c - a|² = 2²→ c² + a² - 2ac = 4Now, DIVIDE the THREE equations by abca²/ABC + b²/abc - 2ab/abc = 1/abc→ a/bc + b/ac - 2/c = 1/60 (since abc = 60)Similarly, you'll GET from the other two equations, b/ac + c/ab - 2/a = 1/60and c/ab + a/bc - 2/b = 4/60Add the three equations, we get2(a/bc + b/ac + c/ab - 1/c - 1/a - 1/b) = 6/60→ a/bc + b/ac + c/ab - 1/c - 1/a - 1/b = 1/20Hence 1/20 is the answer. :) |
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