Привет, помогите пожалуйста решить неравенство:

Question 1

Question 2

$\frac{4^x-2^{x+4}+30}{2^x-2} + \frac{4^x-7*2^{x}+3}{2^x-7} \leq 2^{x+1}-14$
ОДЗ:
$\left \{ {{2^x-2 \neq 0} \atop {2^x-7 \neq 0}} \right.$
$\left \{ {{2^x \neq 2} \atop {2^x \neq 7}} \right.$
$\left \{ {{x \neq 1} \atop {x \neq log_27}} \right.$

$\frac{2^{2x}-16*2^{x}+30}{2^x-2} + \frac{2^{2x}-7*2^{x}+3}{2^x-7} \leq 2*2^{x}-14$
Замена: $2^2=a,$ $a\ \textgreater \ 0$
$\frac{a^2-16a+30}{a-2} + \frac{a^{2}-7a+3}{a-7} \leq 2a-14$
$\frac{(a-7)(a^2-16a+30)}{(a-2)(a-7)} + \frac{(a-2)(a^{2}-7a+3)}{(a-2)(a-7)} \leq \frac{2(a-7)(a-7)(a-2)}{(a-2)(a-7)}$
$\frac{a^3-7a^2-16a^2+112a+30a-210}{(a-2)(a-7)} + \frac{a^3-7a^2+3a-2a^2+14a-6}{(a-2)(a-7)} \leq \frac{2(a-7)(a-7)(a-2)}{(a-2)(a-7)}$
$\frac{a^3-23a^2+142a-210}{(a-2)(a-7)} + \frac{a^3-9a^2+17a-6}{(a-2)(a-7)} \leq \frac{(a^2-14a+49)(2a-4)}{(a-2)(a-7)}$
$\frac{a^3-23a^2+142a-210+a^3-9a^2+17a-6}{(a-2)(a-7)} \leq \frac{a^3-28a^2+98a-4a^2+56a-196}{(a-2)(a-7)}$
$\frac{2a^3-32a^2+159a-216}{(a-2)(a-7)} \leq \frac{2a^3-32a^2+154a-196}{(a-2)(a-7)}$
$\frac{2a^3-32a^2+159a-216}{(a-2)(a-7)} - \frac{2a^3-32a^2+154a-196}{(a-2)(a-7)} \leq 0$
$\frac{2a^3-32a^2+159a-216-(2a^3-32a^2+154a-196)}{(a-2)(a-7)}\leq 0$
$\frac{2a^3-32a^2+159a-216-2a^3+32a^2-154a+196}{(a-2)(a-7)} \leq 0$
$\frac{5a-20}{(a-2)(a-7)} \leq 0$
$\frac{5(a-4)}{(a-2)(a-7)} \leq 0$

--- - ---(2)----+------[4]---- - -----(7)-----+--------
////////// //////////////
---(0)---------------------------------------------------
/////////////////////////////////////////////////////
$0\ \textless \ a\ \textless \ 2$ или $4 \leq a\ \textless \ 7$
$2^x\ \textless \ 2$ или $4 \leq 2^x\ \textless \ 7$
$x\ \textless \ 1$ или $2^2 \leq 2^x\ \textless \ 2^{log_27}$
$2 \leq x\ \textless \ log_27$

Ответ: $(-$ ∞ $;-1)$ ∪ $[2; log_27)$

Question 3

Решение смотрите в приложении.......................

Luluput_zn · Answer 1 · 2018-05-10T07:21:48+0000

$\frac{4^x-2^{x+4}+30}{2^x-2} + \frac{4^x-7*2^{x}+3}{2^x-7} \leq 2^{x+1}-14$
ОДЗ:
$\left \{ {{2^x-2 \neq 0} \atop {2^x-7 \neq 0}} \right.$
$\left \{ {{2^x \neq 2} \atop {2^x \neq 7}} \right.$
$\left \{ {{x \neq 1} \atop {x \neq log_27}} \right.$

$\frac{2^{2x}-16*2^{x}+30}{2^x-2} + \frac{2^{2x}-7*2^{x}+3}{2^x-7} \leq 2*2^{x}-14$
Замена: $2^2=a,$ $a\ \textgreater \ 0$
$\frac{a^2-16a+30}{a-2} + \frac{a^{2}-7a+3}{a-7} \leq 2a-14$
$\frac{(a-7)(a^2-16a+30)}{(a-2)(a-7)} + \frac{(a-2)(a^{2}-7a+3)}{(a-2)(a-7)} \leq \frac{2(a-7)(a-7)(a-2)}{(a-2)(a-7)}$
$\frac{a^3-7a^2-16a^2+112a+30a-210}{(a-2)(a-7)} + \frac{a^3-7a^2+3a-2a^2+14a-6}{(a-2)(a-7)} \leq \frac{2(a-7)(a-7)(a-2)}{(a-2)(a-7)}$
$\frac{a^3-23a^2+142a-210}{(a-2)(a-7)} + \frac{a^3-9a^2+17a-6}{(a-2)(a-7)} \leq \frac{(a^2-14a+49)(2a-4)}{(a-2)(a-7)}$
$\frac{a^3-23a^2+142a-210+a^3-9a^2+17a-6}{(a-2)(a-7)} \leq \frac{a^3-28a^2+98a-4a^2+56a-196}{(a-2)(a-7)}$
$\frac{2a^3-32a^2+159a-216}{(a-2)(a-7)} \leq \frac{2a^3-32a^2+154a-196}{(a-2)(a-7)}$
$\frac{2a^3-32a^2+159a-216}{(a-2)(a-7)} - \frac{2a^3-32a^2+154a-196}{(a-2)(a-7)} \leq 0$
$\frac{2a^3-32a^2+159a-216-(2a^3-32a^2+154a-196)}{(a-2)(a-7)}\leq 0$
$\frac{2a^3-32a^2+159a-216-2a^3+32a^2-154a+196}{(a-2)(a-7)} \leq 0$
$\frac{5a-20}{(a-2)(a-7)} \leq 0$
$\frac{5(a-4)}{(a-2)(a-7)} \leq 0$

--- - ---(2)----+------[4]---- - -----(7)-----+--------
////////// //////////////
---(0)---------------------------------------------------
/////////////////////////////////////////////////////
$0\ \textless \ a\ \textless \ 2$ или $4 \leq a\ \textless \ 7$
$2^x\ \textless \ 2$ или $4 \leq 2^x\ \textless \ 7$
$x\ \textless \ 1$ или $2^2 \leq 2^x\ \textless \ 2^{log_27}$
$2 \leq x\ \textless \ log_27$

Ответ: $(-$ ∞ $;-1)$ ∪ $[2; log_27)$