Ta có : \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2017}\)
\(\Leftrightarrow\frac{bc+ac+ab}{abc}=\frac{1}{a+b+c}\)( do a + b + c = 2017 )
\(\Rightarrow\left(a+b+c\right)\left(bc+ac+ab\right)=abc\)
\(\Leftrightarrow\left(bc+ac\right)\left(a+b+c\right)+ab\left(a+b\right)+abc-abc=0\)
\(\Leftrightarrow c\left(a+b\right)\left(a+b+c\right)+ab\left(a+b\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(ac+bc+c^2+ab\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left[b\left(c+a\right)+c\left(c+a\right)\right]=0\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
Ta có : hoặc a+b =0
hoặc b+c =0
hoặc c+a = 0
Mà \(a+b+c=2017\)
\(\Rightarrow\)hoặc a = 2017; hoặc b = 2017 ; hoặc c = 2017
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