Question

Решите, пожалуйста, 15 задание. У меня почему-то не получается. 50 баллов за ответ

Answer 1

2^x=t
(8t^2-8t+6)/(1-t)+10t+4<=(5-2t^2)/(2-t)
6/(1-t)+2t+4<=2t+4+3/(t-2)
6/(1-t)<=3/(t-2)
2/(1-t)-1/(t-2)<=0
(2(t-2)-(1-t))/((1-t)(t-2)=(3t-5)/((1-t)(t-2)<=0
+++++(1)----[5/3]+++++(2)----
переходя от t к 2^x
++++(0)-----[log(2)(5/3)]+++(1)---
Ответ x=(0;log(2)(5/3))U(1;+беск)

Answer 2

$\frac{2*4^{x+1}-2^{x+3}+6}{1-2^x}+5*2^{x+1}+4 \leq \frac{5-2^{2x+1}}{2-2^x}\\\\ \frac{2*4*2^{2x}-2^3*2^{x}+6}{1-2^x}+5*2*2^{x}+4 \leq \frac{5-2*2^{2x}}{2-2^x}\\\\ 2^x=t\ \textgreater \ 0\\\\ \frac{8t^{2}-8t+6}{1-t}+10t+4 \leq \frac{5-2t^2}{2-t}\\\\ \frac{8t^{2}-8t+6+(10t+4)*(1-t)}{1-t} \leq \frac{5-2t^2}{2-t}\\\\ \frac{8t^{2}-8t+6+10t-10t^2+4-4t}{1-t} \leq \frac{5-2t^2}{2-t}\\\\ \frac{-2t^{2}-2t+10}{1-t} \leq \frac{5-2t^2}{2-t}\\\\ \frac{2t^{2}+2t-10}{t-1} \leq \frac{2t^2-5}{t-2}$

$\frac{(2t^{2}+2t-10)(t-2)-(2t^2-5)(t-1)}{(t-1)(t-2)} \leq 0\\\\ \frac{2t^{3}+2t^2-10t-4t^2-4t+20-[2t^3-5t-2t^2+5]}{(t-1)(t-2)} \leq 0\\\\ \frac{2t^{3}+2t^2-10t-4t^2-4t+20-2t^3+5t+2t^2-5}{(t-1)(t-2)} \leq 0\\\\ \frac{-9t+15}{(t-1)(t-2)} \leq 0\\\\ \frac{t-\frac{5}{3}}{(t-1)(t-2)} \geq 0\\\\ ------(1)++++++[\frac{5}{3}]------(2)+++++\ \textgreater \ t\\\\ t\in(1;\ \frac{5}{3}]\cup(2;\ +\infty)\\\\ ------------------------------\\\\ 1\ \textless \ 2^x \leq \frac{5}{3}\ \ or\ \ 2^x\ \textgreater \ 2$

$2^0\ \textless \ 2^x \leq 2^{log_2(\frac{5}{3})}\ \ or\ \ 2^x\ \textgreater \ 2^1\\\\ 0\ \textless \ x \leq log_2(\frac{5}{3})\ \ or\ \ x\ \textgreater \ 1=log_2(2)\\\\ x\in(0;\ log_2(\frac{5}{3})]\cup(1;\ +\infty)$