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Question 1

Question 2

Решите задачу:

$( \frac{10}{5x-21} + \frac{5x-21}{10} )^2 \leq \frac{25}{4} \; ;\quad ODZ:\quad x\ne \frac{21}{5}\\\\t=\frac{10}{5x-21}\; ,\; \; (t+\frac{1}{t})^2 \leq \frac{25}{4}\; ,\\\\(\frac{t^2+1}{t})^2-(\frac{5}{2})^2 \leq 0\\\\(\frac{t^2+1}{t}-\frac{5}{2})(\frac{t^2+1}{t}+\frac{5}{2}) \leq 0\\\\ \frac{2t^2-5t+2}{2t} \cdot \frac{2t^2+5t+2}{2t} \leq 0\\\\ \frac{2(t-2)(t-\frac{1}{2})\cdot 2(t+2)(t+\frac{1}{2})}{4t^2} \leq 0\quad \quad \frac{(t-2)(t-\frac{1}{2})(t+2)(t+\frac{1}{2})}{t^2} \leq 0$

$+++[-2]---[-\frac{1}{2}]+++(0)+++[\frac{1}{2}]---[2]+++\\\\t \leq -2\; ;\; -\frac{1}{2} \leq t\ \textless \ 0\; ;\; 0\ \textless \ t \leq \frac{1}{2}\; ;\; t \geq 2\\\\ 1)\frac{10}{5x-21} \leq -2\; ;\; \frac{10x-32}{5x-21} \leq 0\; ;\; \quad x\in [3,2\; ;\; 4,2)\\\\ 2)\left \{ {{ \frac{10}{5x-21} \geq - \frac{1}{2} } \atop { \frac{10}{5x-21} \ \textless \ 0}} \right. \; \left \{ {{ \frac{5-1x}{2(5x-21)} \geq 0} \atop {5x-21\ \textless \ 0}} \right. \; \left \{ {{x\in (-\infty ,\;\, 0,2]\cup (4,2\, ;+\infty )} \atop {x\in (-\infty ;\; 4,2)}} \right.$ $x\in (-\infty ;\; 0,2]\\\\3)\; \left \{ {{ \frac{10}{5x-21}\ \textgreater \ 0 } \atop { \frac{10}{5x-21} \leq \frac{1}{2} }} \right. \; \left \{ {{5x-21\ \textgreater \ 0} \atop {\frac{1-5x}{2(5x-21)} \leq 0}} \right. \; \left \{ {{x\ \textgreater \ 4,2} \atop {x\in (-\infty ;\; 0,2]\cup (4,2\; ;+\infty )}} \right. \; \to \; x\in (4,2\; ;\; +\infty )\\\\4)\; \frac{10}{5x-21} \geq 2\; ;\; \frac{52-10x}{5x-21} \geq 0\; ;\; \frac{10x-52}{5x-21} \leq 0\; ;\; x\in (4,2\; ;\; 5,2]\\\\5)\; x\in (-\infty ;\; 0,2]\cup [3,2\; ;\; 4,2)\cup (4,2\; ;5,2]$