Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a+b}{3}=\dfrac{b+c}{4}=\dfrac{c+a}{5}=\dfrac{a+b+b+c+c+a}{3+4+5}=\dfrac{a+b+b+c}{3+4}=\dfrac{b+c+c+a}{4+5}=\dfrac{a+b+c+a}{3+5}=\dfrac{a+b+c}{6}=\dfrac{a+2b+c}{7}=\dfrac{b+2c+a}{9}=\dfrac{b+2a+c}{8}\)
Tiếp tục áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a+b+c}{6}=\dfrac{a+2b+c}{7}=\dfrac{b+2c+a}{9}=\dfrac{b+2a+c}{8}=\dfrac{b+2a+c-a-b-c}{8-6}=\dfrac{a}{2}\) (1)
\(\dfrac{a+b+c}{6}=\dfrac{a+2b+c}{7}=\dfrac{b+2c+a}{9}=\dfrac{b+2a+c}{8}=\dfrac{a+2b+c-a-b-c}{7-6}=\dfrac{b}{1}\)(2)
\(\dfrac{a+b+c}{6}=\dfrac{a+2b+c}{7}=\dfrac{b+2c+a}{9}=\dfrac{b+2a+c}{8}=\dfrac{b+2c+a-a-b-c}{9-6}=\dfrac{c}{3}\) (3)
Từ (1) và (2) và (3) ta có: \(\dfrac{a}{2}=\dfrac{b}{1}=\dfrac{c}{3}\)
Đặt: \(\dfrac{a}{2}=\dfrac{b}{1}=\dfrac{c}{3}=t\Leftrightarrow\left\{{}\begin{matrix}a=2t\\b=t\\c=3t\end{matrix}\right.\)
Ta có: \(M=10a+b-7c+2017=20t+t-21t+2017=21t-21t+2017=0+2017=2017\)Vậy \(M=2017\)