Нужна срочная помощь с Алгеброй, 11 класс!!!. Created by ggfgyyff5655. algebra-ru

решаем методом интервалов____-2____(0)__+___(3)_-_4__+__обратная заменаответ:

Нужна срочная помощь с Алгеброй, 11 класс!!!...

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$\displaystyle \frac{2^{3x}*2+5}{8^x-3}-8^x\geq 3+\frac{6*5^x}{5^x*8^x-3*5^x}\\\\\frac{2*8^x+5}{8^x-3}-8^x\geq 3+\frac{6*5^x}{5^x(8^x-3)}\\\\8^x=t; t>0\\\\\frac{2t+5}{t-3}-t\geq 3+\frac{6}{t-3}\\\\\frac{2t+5}{t-3}-\frac{6}{t-3}-(t+3)\geq 0\\\\\frac{2t-1}{t-3}-\frac{(t+3)(t-3)}{t-3}\geq 0\\\\\frac{2t-1-t^2+9}{t-3}\geq 0\\\\\frac{t^2-2t-8}{t-3}\leq 0$

решаем методом интервалов

$\displaystyle \left \{ {{t^2-2t-8=0} \atop {t-3\neq 0}} \right.; \left \{ {{t=4; t=-2} \atop {t\neq 3}} \right.$

____-2____(0)__+___(3)_-_4__+__

$\displaystyle 3$

обратная замена

$\displaystyle 3$

ответ:

$\displaystyle (\frac{log_23}{3};\frac{2}{3}]$

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Universalka

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$\displaystyle\bf\\\frac{2^{3x+1} +5}{8^{x} -3} -8^{x} \geq 3+\frac{6\cdot 5^{x} }{40^{x}-3\cdot 5^{x} } \\\\\\\frac{2\cdot 8^{x} +5}{8^{x} -3} -8^{x} \geq 3+\frac{6\cdot 5^{x} }{5^{x} \cdot 8^{x}-3\cdot 5^{x} } \\\\\\\frac{2\cdot 8^{x} +5}{8^{x} -3} -8^{x} \geq 3+\frac{6\cdot 5^{x} }{5^{x} (\cdot 8^{x}-3) } \\\\\\\frac{2\cdot 8^{x} +5}{8^{x} -3} -8^{x} \geq 3+\frac{6 }{8^{x}-3} \\\\\\\frac{2\cdot 8^{x} +5}{8^{x} -3} -\frac{6 }{8^{x}-3} -8^{x} -3\geq 0$

$\displaystyle\bf\\\frac{2\cdot 8^{x} -1}{8^{x} -3} -8^{x} -3\geq 0\\\\\\8^{x} =m \ ; \ m>0\\\\\\\frac{2m-1}{m-3} -m-3\geq 0\\\\\\\frac{2m-1-m^{2} -3m+3m+9}{m-3} \geq 0\\\\\\\frac{m^{2} -2m-8}{m-3} \leq 0\\\\\\\frac{(m-4)(m+2)}{m-3} \leq 0$

- - - - - [ - 2] + + + + + (3) - - - - - [4] + + + + + m

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$\displaystyle\bf\\\left \{ {{m>3} \atop {m\leq 4}} \right. \\\\\\\left \{ {{8^{x} >3} \atop {8^{x} \leq 4}} \right. \\\\\\\left \{ {{\log_{8} 8^{x} >\log_{8} 3} \atop {2^{3x} }\leq 2^{2} } \right. \\\\\\\left \{ {{x>\log_{8} 3} \atop {3x\leq 2}} \right. \\\\\\\left \{ {{x>\log_{8}3 } \atop {x\leq \frac{2}{3} }} \right. \\\\\\Otvet:x\in\Big(\log_{8} 3 \ ; \ \frac{2}{3} \Big]$

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