Question

Даю 10 Балов Помогите БАН ЗА ФИГНЮ

Answer 1

$\displaystyle\bf\\1)\\\\5^{3,5+x} \geq \frac{1}{125} \\\\\\5^{3,5+x} \geq 5^{-3} \\\\\\5 > 1 \ \ \Rightarrow \ \ 3,5x\geq -3\\\\\\x\geq -3:3,5\\\\\\x\geq -\frac{6}{7} \\\\\\Otvet \ : \ x\in\Big[-\frac{6}{7} \ ; \ +\infty\Big)\\\\2)\\\\\Big(\frac{1}{7} \Big)^{x} > -1\\\\\\x\in R\\\\Otvet \ : \ x\in(-\infty \ ; \ +\infty)$

$\displaystyle\bf\\3)\\\\\Big(\frac{1}{4} \Big)^{6x-x^{2} } > 4^{-5} \\\\\\\Big(4^{-1} \Big)^{6x-x^{2} } > 4^{-5} \\\\\\4^{x^{2} -6x} > 4^{-5} \\\\\\4 > 1 \ \ \Rightarrow \ \ x^{2} -6x > -5\\\\x^{2} -6x+5 > 0\\\\(x-1)\cdsot(x-5) > 0\\\\\\+ + + + + (1)- - - - - (5)+ + + + + \\\\\\Otvet \ : \ x\in(-\infty \ ; \ 1)\cup(5 \ ; \ +\infty)$