Question

Решите систему неравенств: а){2x-3> 0 б){3-2x< 1 7x+4> 0 1,6+х< 0 найдите целые решения системы неравенств: {6-2x< 3(х-1) 6-два целых икс десятых> x

Answer 1

а) 
 $\left \{ {{2x-3\ \textgreater \ 0} \atop {7x+4\ \textgreater \ 0}} \right. \\ \\ \left \{ {{2x\ \textgreater \ 3} \atop {7x\ \textgreater \ -4}} \right. \\ \\ \left \{ {{x\ \textgreater \ 1.5} \atop {x\ \textgreater \ - \frac{4}{7} }} \right.$x∈(1.5; +∞)

б)
$\left \{ {{3-2x\ \textless \ 0} \atop {1.6+x\ \textless \ 0}} \right. \\ \\ \left \{ {{2x\ \textgreater \ 3} \atop {x\ \textless \ -1.6}} \right. \\ \\ \left \{ {{x\ \textgreater \ 1.5} \atop {x\ \textless \ -1.6}} \right.$

x=∅ у системы неравенств  нет решений   

 $\left \{ {{6-2x\ \textless \ 3(x-1)} \atop {6-2.1x\ \textgreater \ x}} \right. \\ \\ \left \{ {{6-2x\ \textless \ 3x-3)} \atop {6\ \textgreater \ 3.1x}} \right. \\ \\ \left \{ {{5x\ \textgreater \ 3)} \atop {3.1x\ \textless \ 6}} \right. \\ \\ \left \{ {{x\ \textgreater \ 0.6} \atop {x\ \textless \ \frac{60}{31} }} \right.$

x= 1
ответ только одно значение 1