Se $a+\dfrac{1}{b}=\dfrac{7}{3}$, $b+\dfrac{1}{c}=4$ e $c+\dfrac{1}{a}=1$, então qual o valor de $a \cdot b \cdot c$?
Solução:
$a+\dfrac{1}{b}=\dfrac{7}{3}$ (i)
$b+\dfrac{1}{c}=4$ (ii)
$c+\dfrac{1}{a}=1$ (iii)
Resolvendo (i), temos:
$a+\dfrac{1}{b}=\dfrac{7}{3}$ $\Rightarrow$ $\dfrac{ab+1}{b}=\dfrac{7}{3}$ $\Rightarrow$ $7b=3(ab+1)$ $\Rightarrow$ $7b=3ab+3$ $\Rightarrow$ $7b-3ab=3$ $\Rightarrow$ $b(7-3a)=3$ $\Rightarrow$ $b=\dfrac{3}{7-3a}$
Resolvendo (ii), temos:
$b+\dfrac{1}{c}=4$ $\Rightarrow$ $\dfrac{bc+1}{c}=4$ $\Rightarrow$ $bc+1=4c$ $\Rightarrow$ $bc-4c=-1$ $\Rightarrow$ $c(b-4)=-1$ $\Rightarrow$ $c=-\dfrac{1}{(b-4)}$
Resolvendo (iii), temos:
$c+\dfrac{1}{a}=1$ $\Rightarrow$ $\dfrac{ac+1}{a}=1$ $\Rightarrow$ $a=ac+1$ $\Rightarrow$ $a-ac=1$ $\Rightarrow$ $a(1-c)=1$ $\Rightarrow$ $a=\dfrac{1}{1-c}$
Substituindo o resultado de (iii) em (i), temos:
$b=\dfrac{3}{7-3a}$ $\Rightarrow$ $b=\dfrac{3}{7-3\left(\dfrac{1}{1-c}\right)}$ $\Rightarrow$ $b=\dfrac{3}{7-\dfrac{3}{1-c}}$ $\Rightarrow$ $b=\dfrac{3}{\dfrac{7(1-c)-3}{1-c}}$ $\Rightarrow$ $b=\dfrac{3}{\dfrac{7-7c-3}{1-c}}$ $\Rightarrow$ $b=\dfrac{3}{\dfrac{4-7c}{1-c}}$ $\Rightarrow$ $b=3 \cdot \left(\dfrac{1-c}{4-7c}\right)$ $\Rightarrow$ $b=\dfrac{3-3c}{4-7c}$ (iv)
Comparando, agora, (ii) com (iv), temos:
(ii): $b+\dfrac{1}{c}=4$ $\Rightarrow$ $b=4-\dfrac{1}{c}$ $\Rightarrow$ $b=\dfrac{4c-1}{c}$ (v)
Agora, comparando (iv) e (v), temos:
$\dfrac{3-3c}{4-7c}=\dfrac{4c-1}{c}$ $\Rightarrow$ $c(3-3c)=(4c-1)(4-7c)$ $\Rightarrow$ $3c-3c^{2}=16c-28c^{2}-4+7c$ $\Rightarrow$ $25c^{2}-20c+4=0$
$\Delta=b^{2}-4ac$ $\Rightarrow$ $\Delta=(-20)^{2}-4 \cdot 25 \cdot 4$ $\Rightarrow$ $\Delta=400-400$ $\Rightarrow$ $\Delta=0$
$c=\dfrac{-b\pm\sqrt{\Delta}}{2a}$ $\Rightarrow$ $c=\dfrac{20\pm\sqrt{0}}{2 \cdot 25}$ $\Rightarrow$ $c=\dfrac{20\pm0}{50}$ $\Rightarrow$ $c=\dfrac{20}{50}$ $\Rightarrow$ $c=\dfrac{2}{5}$
Para obtermos o valor de $b$, podemos utilizar a igualdade (v):
$b=\dfrac{4c-1}{c}$ $\Rightarrow$ $b=\dfrac{4 \cdot \dfrac{2}{5}-1}{\dfrac{2}{5}}$ $\Rightarrow$ $b=\dfrac{\dfrac{8}{5}-1}{\dfrac{2}{5}}$ $\Rightarrow$ $b=\dfrac{\dfrac{8-5}{5}}{\dfrac{2}{5}}$ $\Rightarrow$ $b=\dfrac{\dfrac{3}{5}}{\dfrac{2}{5}}$ $\Rightarrow$ $b=\dfrac{3}{5} \cdot \dfrac{5}{2}$ $\Rightarrow$ $b=\dfrac{3}{2}$
Escolhendo a igualdade (iii), para obter $a$, temos:
$c+\dfrac{1}{a}=1$ $\Rightarrow$ $\dfrac{2}{5}+\dfrac{1}{a}=1$ $\Rightarrow$ $\dfrac{1}{a}=1-\dfrac{2}{5}$ $\Rightarrow$ $\dfrac{1}{a}=\dfrac{5-2}{5}$ $\Rightarrow$ $\dfrac{1}{a}=\dfrac{3}{5}$ $\Rightarrow$ $3a=5$ $\Rightarrow$ $a=\dfrac{5}{3}$
Como agora temos os valores de $a$, $b$ e $c$, podemos calcular o valor de $a \cdot b \cdot c$. Logo, temos:
$a \cdot b \cdot c=\dfrac{5}{3} \cdot \dfrac{3}{2} \cdot \dfrac{2}{5}=\dfrac{5}{5} \cdot \dfrac{3}{3} \cdot \dfrac{2}{2}=1 \cdot 1 \cdot 1=1$
Assim, temos que $a \cdot b \cdot c=1$.