Đáp án: $M=1$
Giải thích các bước giải:
Ta có:
$\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ca}{c+a}$
$\to \dfrac{a+b}{ab}=\dfrac{b+c}{bc}=\dfrac{c+a}{ca}$
$\to\dfrac1b+\dfrac1a=\dfrac1c+\dfrac1b=\dfrac1a+\dfrac1c$
$\to \dfrac1b+\dfrac1a=\dfrac1c+\dfrac1b\to \dfrac1a=\dfrac1c\to a=c$
Và $\dfrac1c+\dfrac1b=\dfrac1a+\dfrac1c\to \dfrac1b=\dfrac1a\to b=a$
$\to a=b=c$
$\to M=\dfrac{a^2+a^2+a^2}{a^2+a^2+a^2}=1$