Даю 80 баллов Срочно!!!!!!

0 голосов
61 просмотров

Даю 80 баллов Срочно!!!!!!


image

Алгебра (77 баллов) | 61 просмотров
Дан 1 ответ
0 голосов
Правильный ответ

1)\; \; \left\{\begin{array}{l}x-5<7\\x^2-3x+2\geq 0\end{array}\right\; \; \left\{\begin{array}{l}x<12\\(x-1)(x-2)\geq 0\end{array}\right\; \; \left\{\begin{array}{l}x<12\\x\in (-\infty ,1\, ]\cup [\, 2,+\infty )\end{array}\right\\\\\\x\in (-\infty ,1\, ]\cup [\, 2,12)\\\\\\2)\; \; \left\{\begin{array}{l}2x+1\leq x+5\\x^2-8x+12<0\end{array}\right\; \; \left\{\begin{array}{l}x\leq 4\\(x-2)(x-6)<0\end{array}\right\; \; \left\{\begin{array}{l}x\leq 4\\x\in (2\, ;\, 6)\end{array}\right\\\\\\x\in (2\, ;\, 4\, ]

3)\; \; \left\{\begin{array}{l}x+5<7\\x^2-7x+6\leq 0\end{array}\right\; \; \left\{\begin{array}{l}x<2\\(x-1)(x-6)\leq 0\end{array}\right\; \; \left\{\begin{array}{ccc}x<2\\x\in [\; 1\, ;\, 6\; ]\end{array}\right\\\\\\x\in [\; 1\, ;\, 2\, )

image0\end{array}\right\; \; \left\{\begin{array}{l}3x\leq 9\\(x+7)(x-9)>0\end{array}\right\; \; \left\{\begin{array}{l}x\leq 3\\x\in (-\infty ;-7)\cup (9;+\infty )\end{array}\right\\\\\\x\in (-\infty ;-7)" alt="4)\; \; \left\{\begin{array}{l}7-2x\geq x-2\\x^2-2x-63>0\end{array}\right\; \; \left\{\begin{array}{l}3x\leq 9\\(x+7)(x-9)>0\end{array}\right\; \; \left\{\begin{array}{l}x\leq 3\\x\in (-\infty ;-7)\cup (9;+\infty )\end{array}\right\\\\\\x\in (-\infty ;-7)" align="absmiddle" class="latex-formula">

(832k баллов)