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02/08/2017

Sistema não linear (II)

\[ \left\{ {\begin{array}{l} {x + \displaystyle\frac{1}{y} = 1} \\ {y + \displaystyle\frac{1}{z} = 2} \\ {z + \displaystyle\frac{1}{x} = 3} \end{array}} \right. \] Determinar o valor de $xyz$.


Começarei por reescrever o sistema na forma \[ \left\{ {\begin{array}{l} {x + \displaystyle\frac{1}{y} = 1} \\ {y = 2 - \displaystyle\frac{1}{z}} \\ {z = 3 -\displaystyle\frac{1}{x}} \end{array}} \right. \] Assim a primeira equação pode reescrever-se na forma \begin{eqnarray*} {x + \displaystyle\frac{1}{2 - \displaystyle\frac{1}{z}}}&{=}&{1}\\ {\Leftrightarrow x + \frac{1}{2 - \displaystyle\frac{1}{3 -\displaystyle\frac{1}{x}}}}&{=}&{1}\\ {\Leftrightarrow x + \displaystyle\frac{1}{2 - \displaystyle\frac{x}{3x -1}}}&{=}&{1}\\ {\Leftrightarrow x + \displaystyle\frac{3x -1}{5x-2}}&{=}&{1}\\ {\Leftrightarrow 5x^2-2x +3x -1}&{=}&{5x-2}\\ {\Leftrightarrow 5x^2-4x+1}&{=}&{0}\\ {x}&{=}&{\frac{4 \pm \sqrt{16-20}}{2\times 5}}\\ {}&{=}&{\frac{4 \pm 2i}{2\times 5}}\\ {}&{=}&{\frac{2 \pm i}{5}}\\ \end{eqnarray*} Sabendo os possíveis valores de $x$, facilmente se determinam os de $y$ e $z$.
Aqui para poupar escrita, utilizarei a convenção seguinte: nos sinais do tipo $\pm$ as soluções correspondentes ao sinal "de cima" vão corresponder sempre a soluções com o sinal de cima, e o de baixo, aos de baixo.
Portanto: \[ y = \frac{1}{{1 - x}} = \frac{1}{{1 - \frac{{2 \pm i}}{5}}} = \frac{5}{{5 - 2 \mp i}} = \frac{5}{{3 \mp i}} = \frac{{5\left( {3 \pm i} \right)}}{{3^2 + 1^2 }} = \frac{{3 \pm i}}{2} \] e \[ z = 3 - \frac{1}{x} = 3 - \frac{1}{{\displaystyle\frac{{2 \pm i}}{5}}} = 3 - \frac{5}{{2 \pm i}} = 3 - \frac{{5\left( {2 \mp i} \right)}}{{2^2 + 1^2 }} = 3 - 2 \pm i = 1 \pm i \] Agora é só uma questão de calcular o produto... \[xyz=\left(\frac{2 \pm i}{5}\right)\left(\frac{{3 \pm i}}{2}\right)\left(1 \pm i\right)=\left(\frac{1 \pm i}{2}\right)\left(1 \pm i\right)=\frac{\left(1 \pm i\right)^2}{2}=\pm i\]
PS: Uma vez que esta primeira resolução não tem nada de especial, assim que me fôr possível penso noutra e partilho

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