Question

спрочно надо решите пж

Answer 1

$1)\;27^x\leq\left(\frac13\right)^{-6}\\(3^3)^x\leq3^6\\3^{3x}\leq6\\3x\leq6\\x\leq2\\\\2)\;8\cdot2^{x^2-3x}>(0,5)^{-1}\\2^3\cdot2^{x^2-3x}>\left(\frac12\right)^{-1}\\2^{x^2-3x+3}>2^1\\x^2-3x+3>1\\x^2-3x+2>0\\(x-1)(x-2)>0\\x\in(-\infty;\;1)\cup(2;\;+\infty)\\\\3)\;2\log_{0,4}(x-2)\geq\log_{0,4}(x+4)\\O.D.3.:\\\begin{cases}x-2>0\\x+4>0\end{cases}\Rightarrow\begin{cases}x>2\\x>-4\end{cases}\Rightarrow x>2$

$\log_{0,4}(x-2)^2\geq\log_{0,4}(x+4)\\(x-2)^2\leq x+4\\x^2-4x+4\leq x+4\\x^2-5x\leq0\\x(x-5)\leq0\\x\in[0;\;5]\\C\;O.D.3.\;x\in(2;\;5]$