Giải thích các bước giải:
$\dfrac{ab+ac}{2}=\dfrac{bc+ba}{3}+\dfrac{ca+cb}{4}=\dfrac{ab+ac+bc+ba+ca+cb}{2+3+4}$
$\to \dfrac{ab+ac}{2}=\dfrac{bc+ba}{3}+\dfrac{ca+cb}{4}=\dfrac{2(ab+bc+ca)}{9}$
$\to \dfrac{ab+ac}{2}=\dfrac{bc+ba}{3}+\dfrac{ca+cb}{4}=\dfrac{ab+bc+ca}{\dfrac{9}{2}}$
$\to \dfrac{ab+bc+ca-(ab+ac)}{\dfrac{9}{2}-2}=\dfrac{ab+bc+ca-(bc+ba)}{\dfrac{9}{2}-3}=\dfrac{ab+bc+ca-(ca+bc)}{\dfrac{9}{2}-4}$
$\to \dfrac{bc}{\dfrac{5}{2}}=\dfrac{ca}{\dfrac{3}{2}}=\dfrac{ab}{\dfrac{1}{2}}$
$\to \dfrac{bc}{5}=\dfrac{ca}{3}=\dfrac{ab}{1}$
$\to \dfrac{bc}{5}=ab\to \dfrac c5=a, \dfrac{ca}{3}=ab\to b=\dfrac c3\to c=3b$
$\to a=\dfrac c5=\dfrac{3b}{5}$
$\to \dfrac a3=\dfrac b5=\dfrac{c}{15}$
