Question

решите систему ненавенств {2^х +3*2^-х <=4&2х^2-8х/х-7 <=х

Answer 1

$\left \{ {{2^{x}+3\cdot 2^{-x} \leq 4} \atop { \frac{2x^2-8x}{x-7}\leq x }} \right. \\\\1)\; \; 2^{x}+\frac{3}{2^{x}}-4 \leq 0\; ,\; \; t=2^{x}\ \textgreater \ 0\\\\t+ \frac{3}{t} -4 \leq 0\; ,\; \; \frac{t^2-4t+3}{t} \leq 0\; ,\; \; \frac{(t-1)(t-3)}{t} \leq 0\\\\Znaki:\; \; \; ---(0)+++[\, 1\, ] ---[\, 3\, ]+++\\\\t\in (-\infty ,0)\cup [\, 1,3\, ]\\\\t\ \textgreater \ 0\; \; \Rightarrow \; \; 1 \leq 2^{x} \leq 3\; \; \Rightarrow \; \; 0 \leq x \leq log_23$

$2)\; \; \frac{2x^2-8x}{x-7} -x \leq 0\\\\ \frac{2x^2-8x-x^2+7x}{x-7}\leq 0$

$\frac{x^2-x}{x-7} \leq 0\; ,\; \; \; \frac{x(x-1)}{x-7} \leq 0\\\\Znaki:\; \; \; ---[\, 0\, ]+++[\, 1\, ]---(7)+++\\\\x\in (-\infty ,0\, ]\cup [\, 1,7)\\\\3)\; \; \; 1\ \textless \ log_23\ \textless \ 7\; \; \Rightarrow \; \; \; x\in [\, 1,\, log_23\, ]\; \; -\; \; otvet$