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crs12decoder
user (0)
Pe: 2010-10-11T14:09:38+03:00 In: In:

Un sistem omogen

Nu reusesc sa-l fac…

x^2-5xy+2y^2=-1
x^2+5y^2=21

inmultesc prima ecuatie cu 21 si imi da pe undeva pe acolo 105xy si nu cred ca e bine… cum se face?

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1 raspuns

  1. GreatMath
    veteran (III)

    O idee de rezolvare ar fi:

        \[ \begin{array}{l} \left\{ \begin{array}{l} x^2 - 5xy + 2y^2 = - 1\left| { \cdot \left( {21} \right)} \right. \\ x^2 + 5y^2 = 21 \\ \end{array} \right. \Rightarrow \left\{ \begin{array}{l} 21x^2 - 105xy + 42y^2 = - 21\left| { \cdot \left( {\frac{1}{{xy}}} \right)} \right. \\ x^2 + 5y^2 = 21\left| { \cdot \left( {\frac{1}{{xy}}} \right)} \right. \\ \end{array} \right. \\ \left\{ \begin{array}{l} 21\left( {\frac{x}{y}} \right) - 105 + 42\left( {\frac{y}{x}} \right) = \frac{{ - 21}}{{xy}} \\ \left( {\frac{x}{y}} \right) + 5\left( {\frac{y}{x}} \right) = \frac{{21}}{{xy}} \\ \end{array} \right. \\ - - - - - - - - - - - - - - - - - - - - \\ 22\left( {\frac{x}{y}} \right) + 47\left( {\frac{y}{x}} \right) = 105 \\ \left( {\frac{x}{y}} \right) = t \Rightarrow \left( {\frac{y}{x}} \right) = \frac{1}{t} \Rightarrow 22\left( {\frac{x}{y}} \right) + 47\left( {\frac{y}{x}} \right) = 105 \Leftrightarrow 22t + \frac{{47}}{t} = 105 \Leftrightarrow \\ 22t^2 - 105t + 47 = 0 \Rightarrow \left\{ \begin{array}{l} t_1 = \frac{1}{2} \\ t_2 = \frac{{47}}{{11}} \\ \end{array} \right. \\ \end{array} \]

    Mai departe cred ca este simplu..
    Sper ca nu am pierdut câteva solutii pe drum, e bine sa se verifice acest lucru, cum? gândeste-te

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