Прошу, помогите мне, я очень жду вашей помощи!

Question 1

Question 2

Решите задачу:

$\frac{x^2-5x-3}{x^2-5x+3}+\frac{x^2-5x+24}{x^2-5x} \leq 0\\\\t=x^2-5x\; ,\; \; \; \frac{t-3}{t+3}+\frac{t+24}{t} \leq 0\\\\ \frac{(t-3)\cdot t+(t+24)\cdot (t+3)}{t(t+3)} \leq 0 \; \; ,\; \; \frac{t^2-3t+t^2+3t+24t+72}{t(t+3)} \leq 0\\\\ \frac{2t^2+24t+72}{t(t+3)} \leq 0\; \; ,\; \; \; \frac{2(t^2+12t+36)}{t(t+3)} \leq 0\\\\ \frac{2(t+6)^2}{t(t+3)}\leq 0\; ,\; \; t\ne 0\; ,\; \; t\ne -3\\\\+++[-6\, ]+++(-3)---(0)+++\\\\t\in \{-6\}\cup (-3,0)\; \; \Rightarrow \\\\a)\; \; x^2-5x=-6\; ,\; \; x^2-5x+6=0$

$x_1=2\; ,\; \; x_2=3\; \; (teoreva\; Vieta)\\\\b)\; \; -3\ \textless \ t\ \textless \ 0\; \; \Rightarrow \; \; \left \{ {{x^2-5x\ \textless \ 0} \atop {x^2-5x\ \textgreater \ -3}} \right. \; \; \left \{ {{x(x-5)\ \textless \ 0} \atop {x^2-5x+3\ \textgreater \ 0}} \right. \\\\ \left \{ {{x\in (0,5)} \atop {x\in (-\infty , \frac{5-\sqrt{13}}{2})\cup ( \frac{5+\sqrt{13}}{2},+\infty ) }} \right. \; \; \to \; \; x\in (0, \frac{5-\sqrt{13}}{2})\cup ( \frac{5+\sqrt{13}}{2},5)\\\\\star \; \; x^2-5x+3=0\; ,\; \; D=25-4\cdot 3=13\; ,\; x_{1,2}= \frac{5\pm \sqrt{13}}{2} \\\\x_1= \frac{5-\sqrt{13}}{2}\approx 0,7\; \; ;\; \; x_2=\frac{5+\sqrt{13}}{2}\approx 4,3$

$Otvet:\; \; x\in (0,\frac{5-\sqrt{13}}{2})\cup \{2,3\}\cup (\frac{5+\sqrt{13}}{2},5)\; .$