x + 5 = 7 

      x = 7 - 5 

      x = 2

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=0\text{ và: }ab+bc+ac=0\)

=>\(1+\frac{b+c}{a}+1+\frac{a+c}{b}+1+\frac{b+a}{c}=0\text{ và: }\hept{\begin{cases}ab+bc=-ca\\bc+ca=-ab\\ac+ab=-bc\end{cases}}\)

=>\(\frac{ab+ac}{a^2}+\frac{ab+bc}{b^2}+\frac{bc+ac}{c^2}=-3\)

=>\(-\frac{bc}{a^2}+\frac{-ac}{b^2}+\frac{-ab}{c^2}=-3\)

=>\(\frac{bc}{a^2}+\frac{ac}{b^2}+\frac{ab}{c^2}=3\)

10+10=20