Đáp án:
$P = 3$
Giải thích các bước giải:
$\dfrac1a + \dfrac1b +\dfrac1c = 0$
$\to \dfrac{ab + bc + ca}{abc}=0$
$\to ab + bc + ca = 0$
$\to \begin{cases}a(b+ c) + bc = 0\\b(c + a) + ca =0\\c(a + b) + ab = 0\end{cases}$
$\to \begin{cases}\dfrac{bc}{a} = - b - c\\\dfrac{ca}{b}= - c - a\\\dfrac{ab}{c}=- a - b\end{cases}$
$\to \begin{cases}\dfrac{bc}{a^2}=-\dfrac ba - \dfrac ca\\\dfrac{ca}{b^2}=-\dfrac cb -\dfrac ab\\\dfrac{ab}{c^2}=-\dfrac ac -\dfrac bc\end{cases}$
Ta được:
$P =\dfrac{bc}{a^2} + \dfrac{ca}{b^2} + \dfrac{ab}{c^2}$
$\to P = -\dfrac ba -\dfrac ca -\dfrac cb -\dfrac ab -\dfrac ac -\dfrac bc$
$\to P = -a\left(\dfrac1b +\dfrac1c\right) - b\left(\dfrac1a +\dfrac1c\right) -c\left(\dfrac1a +\dfrac1b\right)$
$\to P = -a\cdot\left(-\dfrac1a\right) - b\cdot\left(-\dfrac1b\right)-c\cdot\left(-\dfrac1c\right)$
$\to P = 1 + 1 + 1 = 3$
