Respuesta:
18
Explicación paso a paso:
[tex]x=\frac{-b+-\sqrt{b^2-4ac} }{2a}\\x=\frac{\sqrt{5 }+-\sqrt{5-4} }{2}\\x=\frac{\sqrt{5}+-1}{2}\\x_1=\frac{\sqrt{5}+1 }{2},x_2=\frac{\sqrt{5}-11 }{2}[/tex]
Reemplazo [tex]x_1[/tex]
[tex](\frac{\sqrt{5}+1 }{2})^6+\frac{1}{(\frac{\sqrt{5}+1 }{2})^6}=\frac{(\sqrt{5}+1)^6}{2^6}+\frac{2^6}{(\sqrt{5}+1)^6}=18[/tex]
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Respuesta:
18
Explicación paso a paso:
[tex]x=\frac{-b+-\sqrt{b^2-4ac} }{2a}\\x=\frac{\sqrt{5 }+-\sqrt{5-4} }{2}\\x=\frac{\sqrt{5}+-1}{2}\\x_1=\frac{\sqrt{5}+1 }{2},x_2=\frac{\sqrt{5}-11 }{2}[/tex]
Reemplazo [tex]x_1[/tex]
[tex](\frac{\sqrt{5}+1 }{2})^6+\frac{1}{(\frac{\sqrt{5}+1 }{2})^6}=\frac{(\sqrt{5}+1)^6}{2^6}+\frac{2^6}{(\sqrt{5}+1)^6}=18[/tex]