Đáp án: $P=a$
Giải thích các bước giải:
Ta có:
$\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ca}{c+a}$
$\to \dfrac{a+b}{ab}=\dfrac{b+c}{bc}=\dfrac{c+a}{ca}$
$\to \dfrac1b+\dfrac1a=\dfrac1c+\dfrac1b=\dfrac1a+\dfrac1c$
$\to \begin{cases}\dfrac1b+\dfrac1a=\dfrac1c+\dfrac1b\\ \dfrac1c+\dfrac1b=\dfrac1a+\dfrac1c\end{cases}$
$\to \begin{cases}\dfrac1a=\dfrac1c\\ \dfrac1b=\dfrac1a\end{cases}$
$\to \begin{cases}c=a\\ b=a\end{cases}$
$\to P=\dfrac{a\cdot a^2+a\cdot a^2+a\cdot a^2}{a^2+a^2+a^2}$
$\to P=\dfrac{3a^3}{3a^2}$
$\to P=a$