Geomania.Org Forumları
Üniversite Hazırlık Cebir => Üniversite Hazırlık Cebir => Konuyu başlatan: yasarfaith - Mayıs 10, 2011, 05:10:51 ös
-
güzel soru...
-
(http://latex.codecogs.com/gif.latex?$\textbf{Let%20$\mathbf{\frac{x^2}{y^2}=a\Leftrightarrow%20\frac{x^8}{y^8}=\frac{a^4}{1}\Leftrightarrow%20\boxed{\mathbf{\frac{x^8+y^8}{x^8-y^8}=\frac{a^4+1}{a^4-1}}$}}}$\\\\\\\%20$\textbf{Then%20$\mathbf{\frac{x^2+y^2}{x^2-y^2}+\frac{x^2-y^2}{x^2+y^2}=k}$}$\\\\\\\%20$\textbf{Convert%20into%20$\mathbf{\frac{a+1}{a-1}+\frac{a-1}{a+1}=k}$%20\Leftrightarrow%20\boxed{\displaystyle\mathbf{a^2=\frac{k+2}{k-2}}}}$\\\\\\\%20$\textbf{Now%20$\mathbf{\frac{x^8+y^8}{x^8-y^8}+\frac{x^8-y^8}{x^8+y^8}=\frac{a^4+1}{a^4-1}+\frac{a^4-1}{a^4+1}=2\times\frac{(k+2)^4+(k-2)^4}{(k+2)^4-(k-2)^4}$}}$)