Giải thích các bước giải:

Ta có:

$\dfrac{ab+ac}3=\dfrac{ab+bc}4=\dfrac{bc+ca}5=\dfrac{ab+ac+ab+bc+bc+ca}{3+4+5}=\dfrac{2(ab+bc+ca)}{12}=\dfrac{ab+bc+ca}6$

$\to \dfrac{ab+bc+ca-(ab+ac)}{6-3}=\dfrac{ab+bc+ca-(ab+bc)}{6-4}+\dfrac{ab+bc+ca-(bc+ca)}{6-5}$

$\to \dfrac{bc}3=\dfrac{ac}2=\dfrac{ab}1$

$\to \dfrac{bc}3:(abc)=\dfrac{ac}2:(abc)=\dfrac{ab}1:(abc)$

$\to \dfrac1{3a}=\dfrac1{2b}=\dfrac1{c}$

$\to 3a=2b=c$