Question

Решите систему неравенств 2<=|2x+3|<=7
x^2>=1

Answer 1

$\displaystyle\mathtt{\left\{{{2\leq|2x+3|\leq7}\atop{x^2\geq1}}\right\left\{{{\left\{{{|2x+3|\geq2}\atop{|2x+3|\leq7}}\right}\atop{x^2-1\geq0}}\right\left\{{{\left\{{{\left[\begin{array}{ccc}\mathtt{2x+3\leq-2}\\\mathtt{2x+3\geq2}\end{array}\right}\atop{\left[\begin{array}{ccc}\mathtt{2x+3\geq-7}\\\mathtt{2x+3\leq7}\end{array}\right}}\right}\atop{x\in(-\infty;-1]U[1;+\infty)}}\right}$

$\displaystyle\mathtt{\left\{{{\left\{{{\left[\begin{array}{ccc}\mathtt{2x\leq-5}\\\mathtt{2x\geq-1}\end{array}\right}\atop{\left[\begin{array}{ccc}\mathtt{2x\geq-10}\\\mathtt{2x\leq4}\end{array}\right}}\right}\atop{x\in(-\infty;-1]U[1;+\infty)}}\right\left\{{{\left\{{{\left[\begin{array}{ccc}\mathtt{x\leq-\frac{5}{2}}\\\mathtt{x\geq-\frac{1}{2}}\end{array}\right}\atop{\left[\begin{array}{ccc}\mathtt{x\geq-5}\\\mathtt{x\leq2}\end{array}\right}}\right}\atop{x\in(-\infty;-1]U[1;+\infty)}}\right}$

$\displaystyle\mathtt{\left\{{{\left\{{{x\in(-\infty;-\frac{5}{2}]U[-\frac{1}{2};+\infty)}\atop{x\in[-5;2]}}\right}\atop{x\in(-\infty;-1]U[1;+\infty)}}\right\left\{{{x\in[-5;-\frac{5}{2}]U[-\frac{1}{2};2]}\atop{x\in(-\infty;-1]U[1;+\infty)}}\right}$

ОТВЕТ: $\mathtt{x\in[-5;-\frac{5}{2}]U[1;2]}$