Помогите пожалуйста срочно решить 3,4,5 заранее спасибо!!!

Question 1

Question 2

Решите задачу:

$3)\; \; \frac{x^2-3x+24}{x^2-3x+3}\ \textless \ 4\\\\ \frac{x^2-3x+24-4x^2+12x-12}{x^2-3x+3}\ \textless \ 0\\\\\frac{3x^2-9x-12}{x^2-3x+3}\ \textgreater \ 0\\\\ \frac{x^2-3x-4}{x^2-3x+3}\ \textgreater \ 0\\\\\frac{(x+1)(x-4)}{x^2-3x+3}\ \textgreater \ 0\\\\x^2-3x+3\ \textgreater \ 0\; \; pri\; \; x\in R\; ,\; t.k.\; \; D=9-12=-3\ \textless \ 0\; \; \to \\\\(x+1)(x-4)\ \textgreater \ 0\quad +++(-1)---(4)+++\\\\x\in (-\infty ,-1)\cup (4,+\infty )$

$4)\; \; 0,2^{ \frac{2x-3}{x-2}}\ \textgreater \ 5\; \; ,\; \; 0,2= \frac{2}{10}=\frac{1}{5}=5^{-1} \\\\5^{-\frac{2x-3}{x-2}}\ \textgreater \ 5^1\\\\- \frac{2x-3}{x-2}\ \textgreater \ 1\\\\\frac{-2x+3-x+2}{x-2} \ \textgreater \ 0\\\\ \frac{-3x+5}{x-2}\ \textgreater \ 0\\\\\frac{3x-5}{x-2}\ \textless \ 0\qquad +++( \frac{5}{3})---(2)+++\\\\x\in (1\frac{2}{3}\, ,\, 2)$

$5)\; \; 5^{2x+1}\ \textgreater \ 5^{x}+4\\\\5\cdot 5^{2x}-5^{x}-4\ \textgreater \ 0\\\\\underline {t=5^{x}\ \textgreater \ 0}\; ,\; \; \; 5t^2-t-4\ \textgreater \ 0\\\\D=81\; ,\; \; t_= \frac{1-9}{10}=-\frac{4}{5}\ \textless \ 0\; ,\; \; t_2=\frac{1+9}{10} =1\ \textgreater \ 0\\\\+++(-\frac{4}{5})---(0)---(1)+++\\\\t\in (1,+\infty )\\\\5^{x}\ \textgreater \ 1\; \; \Rightarrow \; \; 5^{x}\ \textgreater \ 5^0\; ,\; \; \underline {x\ \textgreater \ 0}$