ta có (ab+ac)/2 = (ba+bc)/3 = (ca+cb)/4 

=ab+ac-ba-bc+ca+cb/2-3+4 = 2ac/3

=ab+ac+ba+bc-ca-cb/2+3-4 = 2ab

=ab+ac-ba-bc-ca-cb/2-3-4 = 2bc/5

=> 2ac/3=2ab=2bc/5

Ta có 2ac/3=2ab/1 =>c/3 = b/1 => c/15 = b/5    (1)

          2ac/3 = 2bc/5 => a/3 = b/5                         (2)

từ (1) và(2) => a/3 = b/5 = c/15

bạn 2-3-4=5 ??

Theo bài ra, ta có: \(\dfrac{ab+ac}{2}=\dfrac{ba+bc}{3}=\dfrac{ca+cb}{4}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(+)\dfrac{ab+ac}{2}=\dfrac{ba+bc}{3}=\dfrac{ca+cb}{4}=\dfrac{ab+ac+ba+bc-ac-cb}{2+3-4}=\dfrac{2ab}{1}\)

\(+)\dfrac{ab+ac}{2}=\dfrac{ba+bc}{3}=\dfrac{ca+cb}{4}=\dfrac{ab+ac-ba-bc+ca+cb}{2-3+4}=\dfrac{2ac}{3}\)

\(+)\dfrac{ab+ac}{2}=\dfrac{ba+bc}{3}=\dfrac{ca+cb}{4}=\dfrac{-ab-ac+ba+bc+ca+cb}{-2+3+4}=\dfrac{2bc}{5}\)

Suy ra:

\(+)\dfrac{2ab}{1}=\dfrac{2ac}{3}\Rightarrow\dfrac{b}{1}=\dfrac{c}{3}\Rightarrow\dfrac{b}{5}=\dfrac{c}{15}\left(1\right)\)

\(+)\dfrac{2ac}{3}=\dfrac{2bc}{5}\Rightarrow\dfrac{a}{3}=\dfrac{b}{5}\left(2\right)\)

Từ (1) và (2) suy ra \(\dfrac{a}{3}=\dfrac{b}{5}=\dfrac{c}{15}\)

Vậy \(\dfrac{a}{3}=\dfrac{b}{5}=\dfrac{c}{15}\)