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