Question

Cho các số a,b,c>0 và \(\dfrac{a+b}{3}=\dfrac{b+c}{4}=\dfrac{c+a}{5}\)

Tính GTBT \(M=10a+b-7c+2017\)

Answer 1

Đặt \(\dfrac{a+b}{3}=\dfrac{b+c}{4}=\dfrac{c+a}{5}=k\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=3k\\b+c=4k\\c+a=5k\end{matrix}\right.\Rightarrow2\cdot\left(a+b+c\right)=12k\Rightarrow a+b+c=6k\)

\(\Rightarrow\left\{{}\begin{matrix}c=3k\left(1\right)\\a=2k\left(2\right)\\b=k\left(3\right)\end{matrix}\right.\)

Thay \(\left(1\right),\left(2\right),\left(3\right)\) vào BT ta có:

\(M=10\cdot2k+k-7\cdot3k+2017\)

\(M=20k+k-21k+2017\)

\(M=21k-21k+2017\)

\(M=2017\)

Vậy \(M=2017\)