Question

1017. Решите неравенства:
1) |9 – х| = 2;
5) |x – 5| <11;
6) |6 – x| > 7.
ДАЮ 20 БАЛЛОВ​

Answer 1

$\displaytyle \tt 1) \: |9-x|=2\\\\ \displaytyle \tt 9-x=2\\\displaytyle \tt 9-x=-2\\\\\displaytyle \tt -x=-7\\\displaytyle \tt -x=-11\\\\ \displaytyle \tt \boxed{\bold{x_1=7}}\\\displaytyle \tt \boxed{\bold{x_2=11}}$

$\displaytyle \tt 5)\: |x-5|<11\\\\\displaytyle \tt x-5<11, \: \: x-5\geq 0\\\displaytyle \tt -(x-5)<11, \: \: x-5<0\\\\\displaytyle \tt x<16, \: \: x\geq 5\\\displaytyle \tt x>-6, \: \: x<5\\\\\displaytyle \tt x\in [5;\:16)\\\displaytyle \tt x\in (-6;\:5)\\\\\displaytyle \tt \boxed{\bold{x\in(-6;\:16)}}$

$\displaytyle \tt 6) \: |6-x|>7\\\\ \displaytyle \tt 6-x>7, \: \: 6-x\geq 0\\\displaytyle \tt -(6-x)>7, \: \: 6-x<0\\\\ \displaytyle \tt x<-1, \: \: x\leq 6\\\displaytyle \tt x>13, \: \: x>6\\\\ \displaytyle \tt x\in(-\infty; \: -1)\\\displaytyle \tt x\in(13; \: +\infty)\\\\ \displaytyle \tt \boxed{\bold{x\in(-\infty; \: -1)\cup (13; \: +\infty)}}$