Question

помогите решить неравенство, плиииз!!!!

Answer 1

$\displaystyle 9(1+5^{1-2x})^{-\frac{1}{2}}-\frac{1}{2}(5+5^{2x})^{\frac{1}{2}}\geq 6^{\frac{1}{2}}*(5^x)^{\frac{1}{2}}\\\\9(1+\frac{5}{5^{2x}})^{-\frac{1}{2}}-\frac{1}{2}(5+5^{2x})^{\frac{1}{2}}\geq \sqrt{6}*(5^x)^{\frac{1}{2}}\\\\9(\frac{5^{2x}}{5^{2x}+5})^{\frac{1}{2}}-\frac{1}{2}(5+5^{2x})^{\frac{1}{2}}\geq \sqrt{6}*(5^x)^{\frac{1}{2}}| : (5^x)^{\frac{1}{2}}\\\\9(\frac{5^x}{5^{2x}+5})^{\frac{1}{2}}-\frac{1}{2}(\frac{5+5^{2x}}{5^x})^{\frac{1}{2}}\geq \sqrt{6}$

$\displaystyle(\frac{5^x}{5^{2x}+5})^{\frac{1}{2}}=t$

$\displaystyle 9t-\frac{1}{2t}\geq \sqrt{6}\\\\\frac{18t^2-2\sqrt{6}t-1}{2t}\geq 0\\\\18t^2-2\sqrt{6}t-1=0\\\\D=4*6+4*18=96=(4\sqrt{6})^2\\\\t_{1.2}=\frac{2\sqrt{6}\pm 4\sqrt{6}}{36}\\\\t_1=\frac{\sqrt{6}}{6}; t_2=-\frac{\sqrt{6}}{18}$

$\displaystyle \frac{18(t-\frac{\sqrt{6}}{6})(t+\frac{\sqrt{6}}{18})}{2t}\geq 0$

решаем методом интервалов

_______ -√6/18________0_______√6/6_______

     <0                         >0             <0                       >0

$\displaystyle -\frac{\sqrt{6}}{18}\leq t<0; t\geq \frac{\sqrt{6}}{6}$

делаем обратную замену

$\displaystyle -\frac{\sqrt{6}}{18}\leq (\frac{5^x}{5^{2x}+5})^{\frac{1}{2}}<0$

решений нет

$\displaystyle (\frac{5^x}{5^{2x}+5})^\frac{1}{2}\geq\frac{\sqrt{6}}{6}\\\\\frac{5^x}{5^{2x}+5}\geq \frac{1}{6}\\\\\frac{6*5^x-5^{2x}-5}{6(5^{2x}+5)}\geq 0\\\\\ 5^x=a\\\\\frac{-(a^2-6a+5)}{6(a^2+5)}\geq 0$

Знаменатель для любых а положителен

$\displaystyle -(a^2-6a+5)\geq 0\\\\-(a-1)(a-5)\geq 0\\\\1\leq a\leq 5\\\\1\leq 5^x\leq 5\\\\0\leq x\leq 1$

Ответ [0;1]