》,《 - больше либо равно решите неравенства : x> 10 11+1/9z ≥5 35-5x< 0 60 ≥5-13x 9,5-11《11,5x+3 29x+4 1/4 ≥33x+3 1/3 49-3(3-2z)《1-4z -0,5(8x+9)-0,9> 4x-3

Ответ:

Dianochka20013089

05.10.2020 15:32

1) $\left \{ {{x\ \textgreater \ 10} \atop {11+\frac{1}{9}x \geq 5 }} \right.$
$\left \{ {{x\ \textgreater \ 10} \atop {9*11+9*\frac{1}{9}x \geq 9*5 }} \right.$
$\left \{ {{x\ \textgreater \ 10} \atop {99+x \geq 45}} \right.$
$\left \{ {{x\ \textgreater \ 10} \atop {x \geq 45-99}} \right.$
$\left \{ {{x\ \textgreater \ 10} \atop {x \geq -54}} \right.$
x∈(10, +∞)
2) $\left \{ {{35-5x\ \textless \ 0} \atop {60 \geq 5-13x}} \right.$
   $\left \{ {{-5x\ \textless \ -35} \atop {13x \geq 5-60}} \right.$
   $\left \{ {{x\ \textgreater \ 7} \atop {13x \geq -55}} \right.$
   $\left \{ {{x\ \textgreater \ 7} \atop {x \geq - \frac{55}{13}}} \right.$
x∈(7, +∞)
3) $\left \{ {{9.5-11 \leq 11.5x+3} \atop {29x+4\frac{1}{4} \geq 33x+3\frac{1}{3}}} \right.$
   $\left \{ {{-1.5 \leq 11.5x+3} \atop {29x+\frac{17}{4} \geq 33x+\frac{10}{3}}} \right.$
   $\left \{ {{-11.5x \leq 3+1.5} \atop {29x-33x \geq \frac{10}{3}-\frac{17}{4} }} \right.$
   $\left \{ {{-11.5x \leq 4.5} \atop {-4x \geq \frac{40-51}{12} \right.$
   $\left \{ {{x \geq -4.5:11.5} \atop {-4x \geq -\frac{11}{12} \right.$
   $\left \{ {{x \geq - \frac{45}{10}:\frac{115}{10}} \atop {x\leq\frac{11}{48}} \right.$
   $\left \{ {{x \geq - \frac{45}{10}*\frac{10}{115}} \atop {x\leq\frac{11}{48}} \right.$
   $\left \{ {{x \geq - 9*\frac{1}{23}} \atop {x\leq\frac{11}{48}} \right.$
   $\left \{ {{x \geq -\frac{9}{23}} \atop {x\leq\frac{11}{48}} \right.$
  x∈ $[-\frac{9}{23},$ $\frac{11}{48}]$