Question

Решите систему неравенств.Помогите очень нужно!!!!!

Answer 1

1) $\left \{ {{ \frac{x}{3}+ \frac{1}{4}(x-2)\ \textgreater \ x- \frac{x-1}{2} } \atop {\frac{0,7x-8,7}{3}- \frac{5}{6}\ \textless \ 0 } } \right.$
a. $\frac{x}{3}+ \frac{1}{4}(x-2)\ \textgreater \ x- \frac{x-1}{2};$
$4x+ 3(x-2)\ \textgreater \ 12x-6(x-1); 4x+3x-6\ \textgreater \ 12x-6x+6;$
$x\ \textgreater \ 12$
b. $\frac{0,7x-8,7}{3}- \frac{5}{6} \ \textless \ 0; 2(0,7x-8,7)- 5\ \textless \ 0; 1,4x\ \textless \ 22,4; x\ \textless \ 16$
решение системы:
$\left \{ {{x\ \textgreater \ 12} \atop {x\ \textless \ 16}} \right. ;$
x∈(12;16)
ответ: (12;16)

2) $\left \{ {{ x-\frac{x+5}{3}- \frac{x-2}{6}\ \textgreater \ \frac{3x+2}{8} } \atop {\frac{3+x}{4} \ \textgreater \ \frac{x-0,2}{2}- \frac{x}{5} } } \right. ;$
a. $x-\frac{x+5}{3}- \frac{x-2}{6}\ \textgreater \ \frac{3x+2}{8}; 24x-8(x+5)- 4(x-2)\ \textgreater \ 3(3x+2);$ $24x-8x-40- 4x+8\ \textgreater \ 9x+6; 3x\ \textgreater \ 38; x\ \textgreater \ 12\frac{2}{3}$
b. $5(3+x) \ \textgreater \ 10(x-0,2)- 4x; 15+5x\ \textgreater \ 10x-2-4x; 17\ \textgreater \ x; x\ \textless \ 17$
решение системы:
$\left \{ {{x\ \textgreater \ 12 \frac{2}{3} } \atop {x\ \textless \ 17}} \right$
x∈(12$\frac{2}{3}$;17)
ответ: x∈(12$\frac{2}{3}$;17)