Question

Допоможіть вирішити даю 45 балів

Answer 1

$\displaystyle\bf\\1)\\\\2^{x} =\sqrt{2} \\\\\\2^{x} =2^{0,5} \\\\\\\boxed{x=0,5}\\\\2)\\\\\Big(\frac{1}{2} \Big)^{x-2} \geq 4^{-x} \\\\\\\Big(2^{-1}\Big)^{x-2} \geq \Big(2^{2} \Big)^{-x} \\\\\\2^{2-x} \geq 2^{-2x} \\\\\\2 > 1 \ \ \Rightarrow \ \ 2-x\geq -2x\\\\\\-x+2x\geq -2\\\\\\x\geq -2\\\\\\\boxed{x\in\Big[-2 \ ; \ +\infty\Big)}\\\\3)\\\\4^{x+3} =\Big(\frac{1}{4}\Big)^{3x-1} \\\\\\4^{x+3} =\Big(4^{-1} \Big)^{3x-1} \\\\\\4^{x+3} =4^{1-3x} }$

$\displaystyle\bf\\x+3=1-3x\\\\\\x+3x=1-3\\\\\\4x=-2\\\\\\\boxed{x=-0,5}$