
ОДЗ:
0\\4-x^{2}>0\\\end{cases}" alt="\begin{cases} 6|x|-3>0\\4-x^{2}>0\\\end{cases}" align="absmiddle" class="latex-formula">
3\\(2-x)(2+x)>0\\\end{cases}" alt="\begin{cases} 6|x|>3\\(2-x)(2+x)>0\\\end{cases}" align="absmiddle" class="latex-formula">
0,5\\-2' alt='\begin{cases} |x|>0,5\\-2' align="absmiddle" class="latex-formula">
0,5\\-2' alt='\begin{cases} x<-0,5, x>0,5\\-2' align="absmiddle" class="latex-formula">

-убывающая,
поэтому 

1) x<0 <img src="https://tex.z-dn.net/?f=x%5E%7B2%7D-6x-7%5Cgeq0" id="TexFormula10" title="x^{2}-6x-7\geq0" alt="x^{2}-6x-7\geq0" align="absmiddle" class="latex-formula">


![(-\infty;-1]\cup[7;+\infty)](https://tex.z-dn.net/?f=%28-%5Cinfty%3B-1%5D%5Ccup%5B7%3B%2B%5Cinfty%29)
при x<0 ответом будет (-2;-1]</p>
2) x>0 


![(-\infty;-7]\cup[1;+\infty)](https://tex.z-dn.net/?f=%28-%5Cinfty%3B-7%5D%5Ccup%5B1%3B%2B%5Cinfty%29)
при x>0 ответом будет [1;2)
Ответ: ![(-2;-1]\cup[1;2)](https://tex.z-dn.net/?f=%28-2%3B-1%5D%5Ccup%5B1%3B2%29)