Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{a+c-b}{b}=\frac{a+b-c+b+c-a+a+c-b}{c+a+b}=\frac{a+b+c}{c+a+b}=1\)
Do đó: \(\frac{a+b-c}{c}=1\)\(\Rightarrow a+b-c=c\)\(\Rightarrow a+b+c=3c\) (1)
\(\frac{b+c-a}{a}=1\)\(\Rightarrow b+c-a=a\)\(\Rightarrow b+c+a=3a\) (2)
\(\frac{a+c-b}{b}=1\)\(\Rightarrow a+c-b=b\)\(\Rightarrow a+c+b=3b\) (3)
Từ (1), (2), (3) \(\Rightarrow3a=3b=3c\)\(\Rightarrow a=b=c\)
Ta có: \(T=\left(10+\frac{b}{a}\right)\left(4+\frac{2c}{b}\right)\left(2017+\frac{3a}{c}\right)\)
\(=\left(10+\frac{a}{a}\right)\left(4+\frac{2c}{c}\right)\left(2017+\frac{3a}{a}\right)\)
\(=\left(10+1\right)\left(4+2\right)\left(2017+3\right)\)
\(=11.6.2020=133320\)
p/s: làm thế này đúng không ta, mình hong chắc lắm