Прошуууууууууууууууууууууууууу Помогите решить систему log2x+1(4x-5)+log4x-5(2x+1)меньше...

Question 1

Прошуууууууууууууууууууууууууу
Помогите решить систему
log2x+1(4x-5)+log4x-5(2x+1)меньше или равно 2
9^x-2*6^x-3*4^x меньше или равно 0

Question 2

Решение 1 неравенства x∈(1,25;1,5)
Решение 2 неравенства x∈(-∞;log(3/2)3}
Общее решение системы x∈(1,25;1,5)

Question 3

$2x+1 \neq 0 \ \Rightarrow 2x =-1 \ \Rightarrow \ x =-\frac{1}{2} \\ 2x+1\ \textgreater \ 0 \ \Rightarrow \ x\ \textgreater \ =-\frac{1}{2} \\ 4x-5 \neq 0 \ \Rightarrow \ 4x \neq 5 \ \Rightarrow \ x \neq \frac{5}{4} \\ 4x-5\ \textgreater \ 0 \ \Rightarrow \ x \ \textgreater \ \frac{5}{4} \\ \\ \\ \log_{2x+1}(4x-5) + \log_{4x-5}(2x+1) \leq 2 \\ \\ \log_{2x+1}(4x-5) + \frac{1}{\log_{2x+1}(4x-5)} -2 \leq 0 \\ \\ \frac{\log^2_{2x+1}(4x-5)-2 \log_{2x+1} (4x-5)+1}{\log_{2x+1}(4x-5)} \leq 0$

$t = \log_{2x+1} (4x-5) \\ \\ \frac{t^2-2t+1}{t} \leq 0; \ \ t \neq 0 \\ \\ (t-1)^2 =0; \ \ t_{1,2}=1$

   — + ч +
----------o------------*------------->x
      0 1

$t\ \textless \ 0; \ \ \ t=1 \\ \\ \\ \log_{2x+1} (4x-5) \ \textless \ 0; \ \ \ (0\ \textless \ 2x+1\ \textless \ 1; \ \ \ -1\ \textless \ 2x\ \textless \ 0; \ \ \ -\frac{1}{2}\ \textless \ x\ \textless \ 0) \\ \\ \log_{2x+1} (4x-5) \ \textless \ \log_{2x+1} 1; \ \ \ 4x-5\ \textgreater \ 1; \ \ 4x \ \textgreater \ 6; \ \ x \ \textgreater \ \frac{3}{2} \ \notin \ [-\frac{1}{2};0] \\ \\ \\ 2x+1\ \textgreater \ 1; \ \ \ 2x\ \textgreater \ 0; \ \ x\ \textgreater \ 0 \\ \log_{2x+1} (4x-5) \ \textless \ \log_{2x+1} 1; \ \ \ 4x-5 \ \textless \ 1; \ \ \ \ 4x \ \textless \ 6; \ \ x \ \textless \ \frac{3}{2} \\ \\ 0 \ \textless \ x \ \textless \ \frac{3}{2} \ \Rightarrow \ \boxed{ \frac{5}{4}\ \textless \ x \ \textless \ \frac{3}{2}}$

$\log_{2x+1} (4x-5)=1; \\ \log_{2x+1} (4x-5)=\log_{2x+1}(2x+1) \\ \\ 4x-5=2x+1 \\ 4x-2x=1+5 \\ 2x=6 \\ \\ \boxed{x=3}$

$9^x-2 \cdot 6^x -3 \cdot 4^x \leq 0 \\ \\ 3^{2x} - 2 \cdot 2^x \cdot 3^x -3 \cdot 2^{2x} \leq 0 \ \ \ \ (a^2 - 2ab-3b^2=(a+b) \cdot (a-3b)); \\ \\ (3^x +2^x) \cdot (3^x - 3 \cdot 2^x) \leq0 \\ \\ 3^x+2^x=0; \ \ 2^x \ \textgreater \ 0, \ 3^x \ \textgreater \ 0 \ \Rightarrow 3^x + 2^x \neq 0 \\ \\ 3^x -3 \cdot 2^x =0; \ \ \ 3^x=3 \cdot 2^x \\ \\ \log_3 3^x =\log_3 (3 \cdot 2^x); \ \ x =\log_33+ x\log_32; \\\\ x \cdot (1-\log_32) = 1; \ \ \ x =\frac{1}{1-\log_32}=\frac{1}{\log_33-\log_32}=\frac{1}{\log_3 \frac{3}{2}}=\log_{\frac{3}{2}}3 \\ \\$

   — +
------------------------*------------------>x
   log[3/2, 3]

$x \leq \log_{\frac{3}{2}} 3$

Сравним логарифм с тройкой.
$3\ \ \ \ ; \ \ \ \ \log_{\frac{3}{2}} 3 \\ \\ \log_{\frac{3}{2}}{\frac{3}{2}}^3 \ \ \ \ ; \ \ \ \log_{\frac{3}{2}}3 \\ \\ \frac{27}{8} \ \ \ \ ; \ \ \ \ 3 \\ \\ \frac{27}{8} \ \textgreater \ \frac{24}{8} \\ \\ 3\ \textgreater \ \log_{\frac{3}{2}}3$

Ответ: $\frac{5}{4}\ \textless \ x \ \textless \ \frac{3}{2}$