(f(2010

الملف الشخصي · 12-15-2014, 02:39 PM

السلام عليكم ورحمة الله وبركاته

الملف الشخصي · 12-15-2014, 09:37 PM

$\\ \because \ {\color{Blue} f\left ( x+1 \right ) =\frac{1+f\left ( x \right )}{1-f\left ( x \right )}} \\ \\ \therefore \ {\color{Blue} f\left ( 2 \right )=\frac{1+f\left ( 1\right )}{1-f\left ( 1 \right )}} \\ \\ \therefore \ {\color{Blue}f\left ( 3 \right )=\frac{1+f\left ( 2 \right )}{1-f\left ( 2 \right )} }=\frac{\left (1+\frac{1+f\left ( 1 \right )}{1-f\left ( 1 \right )} \right )}{\left (1-\frac{1+f\left ( 1 \right ) }{1-f\left ( 1 \right )} \right )} .\frac{\left ( 1-f\left ( 1 \right ) \right )}{\left ( 1-f\left ( 1 \right ) \right )}$

$\\ \therefore \ {\color{Blue} f\left ( 3 \right )}=\frac{\left ( 1-f\left ( 1 \right )+1+f\left ( 1 \right ) \right )}{\left (1-f\left ( 1 \right ) -1-f\left ( 1 \right ) \right )}=\frac{2}{-2f\left ( 1 \right )} \\ \\ \therefore {\color{Blue} \ f\left ( 3 \right )}={\color{Blue} \frac{-1}{f\left ( 1 \right )}}$

وبنفس الطريقة يمكن أثبات أن:

$\\ {\color{Blue} f\left ( 5 \right )=\frac{-1}{f\left ( 3 \right )}} \\ \\ \therefore \ \ {\color{Blue} f\left ( 5 \right )=\frac{-1}{\left ( \frac{-1}{f\left ( 1 \right )} \right )}}\rightarrow {\color{Blue} f\left ( 5 \right )=f\left ( 1 \right )}$

وبنفس الطريقة يمكن أثبات أن:

$\\ {\color{Blue} f\left ( 9 \right )=f\left ( 5 \right )} \\ but \ \ {\color{Blue} f\left ( 5 \right )=f\left ( 1 \right )} \\ \therefore \ \ {\color{Blue} f\left ( 1 \right )=f\left ( 5 \right )=f\left ( 9 \right )}$

وبنفس الطريقة يمكن أثبات أن:

$\\ {\color{Blue} f\left ( 1 \right )=f\left ( 5 \right )=f\left ( 9 \right )=f\left ( 13 \right )=f\left ( 17 \right )=...........} \\ \\ \therefore \ \ {\color{Blue} f\left (1+4m \right )=f\left ( 1 \right ) \ \ \forall \ m\in N} \\ \\ \because \ \ {\color{Blue} f\left ( 2009 \right )=f\left ( 1+2008 \right )=f\left ( 1+4\left ( 502 \right ) \right ) \ : \ 502\in N} \\ \\ \therefore {\color{Blue} \ \ f\left ( 2009 \right )=f\left ( 1 \right )} \\ \\ \therefore \ \ {\color{Blue} f\left ( 2010 \right )=\frac{1+f\left ( 2009 \right )}{1-f\left ( 2009 \right )}=\frac{1+f\left ( 1 \right )}{1-f\left ( 1 \right )}=\frac{1+2}{1-2}}$

$\therefore \ \ {\color{Blue} f\left ( 2010 \right )=-3}$

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