Question by Uchiha Sasuke - Discuss with MathYouLike

Uchiha Sasuke

30/11/2017 at 08:56

Given \(\dfrac{a+b+c}{a}=\dfrac{a+b+c}{b}=\dfrac{a+b+c}{c}\), calculate \(A=\dfrac{a}{a+b+c}+\dfrac{b}{b+c+d}+\dfrac{c}{c+d+a}+\dfrac{d}{d+a+b}\).

Fraction

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Huỳnh Anh Phương 13/06/2018 at 02:46

a, b, c, d \(\ne\)0

a,b,c,d=1

\(\dfrac{a+b+c}{a}+\dfrac{a+b+c}{b}+\dfrac{a+b+c}{c}\)

\(=\dfrac{1+1+1}{1}+\dfrac{1+1+1}{1}+\dfrac{1+1+1}{1}=3\)

\(\Rightarrow A=\dfrac{1}{1+1+1}+\dfrac{1}{1+1+1}+\dfrac{1}{1+1+1}+\dfrac{1}{1+1+1}=\dfrac{1}{3}\times4=\dfrac{4}{3}\)
FC Alan Walker 16/03/2018 at 13:38

Where is "d"?

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