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Fantezi Cebir => Analiz-Cebir => Konuyu başlatan: yasarfaith - Temmuz 15, 2011, 10:07:22 ös

Başlık: $(\sqrt{2+\sqrt 3})^x+(\sqrt{2-\sqrt 3})^x=4$ denkleminin reel köklerini bulunuz
Gönderen: yasarfaith - Temmuz 15, 2011, 10:07:22 ös

$(\sqrt{2+\sqrt 3})^x+(\sqrt{2-\sqrt 3})^x=4$ denkleminin reel köklerini bulunuz.

Başlık: Ynt: denklem 26
Gönderen: stuart clark - Temmuz 16, 2011, 10:06:18 öö

(http://latex.codecogs.com/gif.latex?\left(\sqrt{2+\sqrt{3}}\right).\left(\sqrt{2-\sqrt{3}}\right)=1$\\\\\\%20$\left(\sqrt{2-\sqrt{3}}\right)=\frac{1}{\left(\sqrt{2+\sqrt{3}}\right)}$\\\\\\%20$\left(\sqrt{2+\sqrt{3}}\right)^x+\left(\sqrt{2-\sqrt{3}}\right)^x=4$\\\\\\%20Let%20$\left(\sqrt{2+\sqrt{3}}\right)^x=y,$%20Then\\\\%20$y+\frac{1}{y}=4\Leftrightarrow%20y^2-4y+1=0$\\\\%20$y=\frac{4\pm%202\sqrt{3}}{2}$\\\\%20so%20$y=\left(2\pm%20\sqrt{3}\right)$\\\\%20*If%20$y=\left(2+%20\sqrt{3}\right)\Leftrightarrow%20\left(2+%20\sqrt{3}\right)^x=\left(2+%20\sqrt{3}\right)^1\Leftrightarrow%20\boxed{x=+1}$\\\\%20*If%20$y=\left(2-%20\sqrt{3}\right)\Leftrightarrow%20\left(2+%20\sqrt{3}\right)^x=\left(2+%20\sqrt{3}\right)^{-1}\Leftrightarrow%20\boxed{x=-1}$\\\\)

Başlık: Ynt: denklem 26
Gönderen: yasarfaith - Temmuz 16, 2011, 03:26:39 ös

thx...