Question

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

Answer 1

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

$displaystylemathtt{left{{{left{{{left[begin{array}{ccc}mathtt{2xleq-5}\mathtt{2xgeq-1}end{array}right}atop{left[begin{array}{ccc}mathtt{2xgeq-10}\mathtt{2xleq4}end{array}right}}right}atop{xin(-infty;-1]U[1;+infty)}}rightleft{{{left{{{left[begin{array}{ccc}mathtt{xleq-frac{5}{2}}\mathtt{xgeq-frac{1}{2}}end{array}right}atop{left[begin{array}{ccc}mathtt{xgeq-5}\mathtt{xleq2}end{array}right}}right}atop{xin(-infty;-1]U[1;+infty)}}right}$

$displaystylemathtt{left{{{left{{{xin(-infty;-frac{5}{2}]U[-frac{1}{2};+infty)}atop{xin[-5;2]}}right}atop{xin(-infty;-1]U[1;+infty)}}rightleft{{{xin[-5;-frac{5}{2}]U[-frac{1}{2};2]}atop{xin(-infty;-1]U[1;+infty)}}right}$

ОТВЕТ: $mathtt{xin[-5;-frac{5}{2}]U[1;2]}$