Question

Решить неравенство , пожалуйста

Answer 1

$25^{2x^2-0,5}-0,6\cdot 4^{2x^2+0,5}\leq 10^{2x^2}\\\\5^{2(2x^2-0,5)}-0,6\cdot 2^{2(2x^2+0,5)}\leq 10^{2x^2}\\\\5^{4x^2-1}-\frac{3}{5}\cdot 2^{4x^2+1} \leq 10^{2x^2}\\\\5^{4x^2}\cdot \frac{1}{5}-\frac{3}{5}\cdot 2^{4x^2}\cdot 2-10^{2x^2}}\leq 0\; \Big |:(2\cdot 5)^{2x^2}\\\\\Big (\frac{5}{2}\Big )^{2x^2}\cdot \frac{1}{5}-\frac{6}{5}\cdot \Big (\frac{2}{5}\Big )^{2x^2}-1\leq 0\\\\t=\Big (\frac{5}{2}\Big )^{2x^2}>0\; \; \; \; \Rightarrow \; \; \; \; \frac{1}{5}\cdot t-\frac{6}{5\cdot t}-1\leq 0\; ,\; \; \frac{t^2-5t-6}{5\, t}\leq 0$

$\frac{(t-6)(t+1)}{5\, t}\leq 0\; \; ,\; \; \; znaki:\; \; ---[-1\, ]+++(0)---[\, 6\, ]+++\\\\t\in (-\infty ,-1\, ]\cup (0,6\; ]\\\\t>0\; \; \Rightarrow \; \; t\in (0,6\; ]\\\\0<\Big (\frac{5}{2}\Big )^{2x^2}\leq 6\; \; \; \Rightarrow \quad 2x^2\leq log_{\frac{5}{2}}\, 6\; \; ,\; \; x^2\leq \frac{1}{2}\, log_{2,5}\, 6\; ,\; \; x^2-log_{2,5}\, \sqrt6\leq 0\; ,\\\\-\sqrt{log_{2,5}\, \sqrt6}\leq x\leq \sqrt{log_{2,5}\, \sqrt6}\\\\x\in \Big (-\sqrt{log_{2,5}\, \sqrt6}\; ;\; \sqrt{log_{2,5}\, \sqrt6}\, \Big )$