Problema
Se [tex]~\dfrac{1}{a+1}=2[/tex], [tex]~\dfrac{1}{b+2}=3~[/tex] e [tex]~\dfrac{1}{c+3}=6[/tex], calcular [tex]\dfrac{1}{a+b+c}[/tex].
Solução
Temos que
[tex]\quad\quad\quad \dfrac{x}{y}=z \iff zy=x, \forall \ y\ne0[/tex].
Logo:
[tex]\quad\quad\,\,\bullet\,\,\dfrac{1}{a+1}=2\Rightarrow 2(a+1)=2\Rightarrow a= – \dfrac{1}{2}[/tex];
[tex]\quad\quad\,\,\bullet \,\,\dfrac{1}{b+2}= 3\Rightarrow 3(b+2)=1\Rightarrow b= – \dfrac{5}{3}[/tex];
[tex]\quad\quad\,\,\bullet \,\,\dfrac{1}{c+3}=6\Rightarrow 6(c+3)=1\Rightarrow c= – \dfrac{17}{6}[/tex];
de sorte que
[tex]\quad\quad\quad \dfrac{1}{a+b+c}=\dfrac{1}{-\dfrac{1}{2}-\dfrac{5}{3}-\dfrac{17}{6}}= \dfrac{1}{\dfrac{-3-10-17}{6}}[/tex].
Assim, [tex]\boxed{ \dfrac{1}{a+b+c}= -\dfrac{1}{5}}.[/tex]
Solução elaborada pela aluna do PIC-OBMEP Miriam de Cássia Souza.