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Fantezi Cebir => Analiz-Cebir => Konuyu başlatan: yasarfaith - Haziran 05, 2011, 04:26:59 ös

Başlık: $\sqrt[3]{x+1}+\sqrt[3]{3x+1}=\sqrt[3]{x-1}$
Gönderen: yasarfaith - Haziran 05, 2011, 04:26:59 ös

$\sqrt[3]{x+1}+\sqrt[3]{3x+1}=\sqrt[3]{x-1}$ denklemini çözünüz.

Başlık: Ynt: denklem 38
Gönderen: FEYZULLAH UÇAR - Haziran 05, 2011, 11:10:57 ös

....

Başlık: Ynt: denklem 38
Gönderen: stuart clark - Haziran 09, 2011, 05:35:18 öö

(http://latex.codecogs.com/gif.latex?$\textbf{Let%20$\mathbf{x+1=a^3,\hspace{10}%203x+1=b^3,\hspace{10}%20x-1=-c^3}$}$\\\\%20\mathbf{\sqrt[3]{x+1}+\sqrt[3]{3x+1}-\sqrt[3]{x-1}=0}$\\\\%20$\mathbf{a+b+c=0\Leftrightarrow%20a^3+b^3+c^3=3abc}$\\\\%20$\mathbf{x+1+3x+1-x+1=-3\sqrt[3]{x+1}.\sqrt[3]{3x+1}.\sqrt[3]{x-1}}$\\\\%20$\mathbf{3(x+1)=-3\sqrt[3]{x+1}.\sqrt[3]{3x+1}.\sqrt[3]{x-1}}$\\\\%20$\mathbf{(x+1)^3=-(x+1).(3x+1).(x-1)}$\\\\%20$\mathbf{(x+1).\left\{(x+1)^2+(3x+1).(x-1)\right\}=0}$\\\\%20$\mathbf{(x+1).4x^2=0}$\\\\%20$\textbf{\mathbf{x%20=%20-1,x%20=%200}(Not%20Satisfied)}$\\\\%20\textbf{So%20$\boxed{\boxed{\mathbf{x=-1}}}$})