Giải thích các bước giải:
Ta có :
$\dfrac{ab+ac}{2}=\dfrac{bc+ba}{3}=\dfrac{ca+cb}{4}=\dfrac{ab+ac+bc+ba+ca+bc}{2+3+4}=\dfrac{2(ab+bc+ca)}{9}$
$\rightarrow \dfrac{2(ab+ac)}{4}=\dfrac{2(bc+ba)}{6}=\dfrac{2(ca+cb)}{8}=\dfrac{2(ab+bc+ca)}{9}$
$\rightarrow \dfrac{2(ab+bc+ca)-2(ab+ac)}{9-4}=\dfrac{2(ab+bc+ca)-2(bc+ba)}{9-6}=\dfrac{2(ab+bc+ca)-2(ca+cb)}{9-8}$
$\rightarrow \dfrac{2bc}{5}=\dfrac{2ac}{3}=\dfrac{2ab}{1}$
$\rightarrow \dfrac{bc}{5}=\dfrac{ac}{3}=\dfrac{ab}{1}$
$\rightarrow \dfrac{1}{5a}=\dfrac{1}{3b}=\dfrac{1}{c}$ chia cả tử và mẫu cho abc
$\rightarrow 5a=3b=c$
$\rightarrow\dfrac{a}{3}=\dfrac{b}{5}=\dfrac{c}{15}$ chia cho 15
