Question

Решите неравенство. Пожалуйста

Answer 1

Ответ:

Пошаговое объяснение:

$\displaystyle\\(5+2\sqrt{6} )^{2x}-10\cdot\bigg(\frac{1}{5-2\sqrt{6} }\bigg)^x+1\geq 0\\\\\frac{1}{5-2\sqrt{6} }=\frac{1}{5-2\sqrt{6} }\cdot\frac{5+2\sqrt{6} }{5+2\sqrt{6} }=5+2\sqrt{6}\\\\ (5+2\sqrt{6} )^{2x}-10\cdot(5+2\sqrt{6} )^x+1\geq 0\\\\t= (5+2\sqrt{6} )^x\\\\t^2-10t+1\geq 0\\\\D=b^2-4ac=100-4=96=16*6\\\\t=\frac{10\pm4\sqrt{6} }{2} =5\pm2\sqrt{6} \\\\(t-(5+2\sqrt{6} ))(t-(5-2\sqrt{6} ))\geq 0\\\\((5+2\sqrt{6} )^x-(5+2\sqrt{6} )^1)((5+2\sqrt{6} )^x-(5+2\sqrt{6} )^{-1})\geq 0\\\\(x-1)(x-(-1))\geq 0$

$znaki:+++[-1]---[1]+++>x\\\\Otvet:x\in(-\infty;-1]\cup[1;+\infty)$