a2/(b+c) +b2/(c+a) +c2/(a+b)=0 then prove that a/(b+c) +b/(c+a) +c/(a+b) =1

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Aditya Gupta · Answer

Let a/(b+c) +b/(c+a) +c/(a+b)= k Multiply both sides by a+b+c to get [a2/(b+c)+a] + [b2/(c+a)+b] + [c2/(a+b)+c] = k(a+b+c) Or a2/(b+c) +b2/(c+a) +c2/(a+b) + (a+b+c)= k(a+b+c) Or a+b+c= k(a+b+c) Or k= 1 QED KINDLY APPROVE :))

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a2/(b+c) +b2/(c+a) +c2/(a+b)=0 then prove that a/(b+c) +b/(c+a) +c/(a+b) =1

a2/(b+c) +b2/(c+a) +c2/(a+b)=0 then prove that a/(b+c) +b/(c+a) +c/(a+b) =1

a2/(b+c) +b2/(c+a) +c2/(a+b)=0 then prove that a/(b+c) +b/(c+a) +c/(a+b) =1

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a2/(b+c) +b2/(c+a) +c2/(a+b)=0 then prove that a/(b+c) +b/(c+a) +c/(a+b) =1

a2/(b+c) +b2/(c+a) +c2/(a+b)=0 then prove that a/(b+c) +b/(c+a) +c/(a+b) =1

a2/(b+c) +b2/(c+a) +c2/(a+b)=0 then prove that a/(b+c) +b/(c+a) +c/(a+b) =1

1 Answers

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