\(\dfrac{a+b+c}{a+b-c}=\dfrac{a-b+c}{a-b-c}\)
Theo tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{a+b+c}{a+b-c}=\dfrac{a-b+c}{a-b-c}=\dfrac{a+b+c-\left(a-b+c\right)}{a+b-c-\left(a-b-c\right)}\)
\(=\dfrac{a+b+c-a+b-c}{a+b-c-a+b+c}=\dfrac{2b}{2b}=1\)
\(\Rightarrow a+b+c=a+b-c\)
\(\Rightarrow c=-c\)
\(\Rightarrow c-\left(-c\right)=0\)
\(\Rightarrow c+c=0\)
\(\Rightarrow2c=0\)
\(\Rightarrow c=0\)
Vậy \(c=0\)
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