Theo bài ra: \(\frac{a+b}{6}=\frac{b+c}{4}=\frac{c+a}{3}\)
Đặt \(\frac{a+b}{6}=\frac{b+c}{4}=\frac{c+a}{3}=t\left(t\ne0\right)\Rightarrow a+b=6t,b+c=4t,c+a=3t\)
\(\Rightarrow\left(a+b\right)+\left(b+c\right)+\left(c+a\right)=6t+4t+3t\)
\(\Rightarrow2\left(a+b+c\right)=13t\Rightarrow a+b+c=6,5t\)
Ta có: \(a=\left(a+b+c\right)-\left(b+c\right)=6,5t-4t=2,5t\)
\(b=\left(a+b\right)-a=6t-2,5t=3,5t\)
\(c=\left(a+c\right)-a=3t-2,5t=0,5t\)
Ta có: \(b-a=3,5t-2,5t=t\)
\(a-c=2,5t-0,5t=2t\)
\(b-c=3,5t-0,5t=3t\)
Vậy \(\left(b-a\right):\left(a-c\right):\left(b-c\right)=t:\left(2t\right):\left(3t\right)=1:2:3\)
hay b - a , a - c và b - c tỉ lệ thuận với 1,2,3
