Đáp án: $ K = 3$
Giải thích các bước giải:
$ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0 ⇔ \frac{1}{c} = - (\frac{1}{a} + \frac{1}{b}) ⇔ \frac{ab}{c²} = - (\frac{ab}{ca} + \frac{ab}{bc}) (1)$
$ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0 ⇔ \frac{1}{a} = - (\frac{1}{b} + \frac{1}{c}) ⇔ \frac{bc}{a²} = - (\frac{bc}{ab} + \frac{bc}{ca}) (2)$
$ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0 ⇔ \frac{1}{b} = - (\frac{1}{c} + \frac{1}{a}) ⇔ \frac{ca}{b²} = - (\frac{ca}{bc} + \frac{ca}{ab}) (3)$
$(1) + (2) + (3) : $
$ K = \frac{ab}{c²} + \frac{bc}{a²} + \frac{ca}{b²} = - \frac{ab + bc}{ca} - \frac{bc + ca}{ab} - \frac{ca + ab}{bc}$
$ = - b(\frac{1}{c} + \frac{1}{a}) - c(\frac{1}{a} + \frac{1}{b}) - a(\frac{1}{b} + \frac{1}{c})$
$ = - b(- \frac{1}{b}) - c(- \frac{1}{c}) - a(-\frac{1}{a}) = 1 + 1 + 1 = 3$
