• Предмет: Алгебра
  • Автор: alonapolah
  • Вопрос задан 1 год назад

Срочно!!!!!!!!!!!!!! 30 б.​

Приложения:

Ответы

Ответ дал: Universalka
1

\displaystyle\bf\\\left \{ {{3(x+1)+2(x-2)\leq 4x-5} \atop {4(x-2)-3(x+1)\leq 2x-2}} \right. \\\\\\\left \{ {{3x+3+2x-4\leq 4x-5} \atop {4x-8-3x-3\leq 2x-2}} \right. \\\\\\\left \{ {{5x-1\leq 4x-5} \atop {x-11\leq 2x-2}} \right. \\\\\\\left \{ {{5x-4x\leq -5+1} \atop {x-2x\leq -2+11}} \right. \\\\\\\left \{ {{x\leq -4} \atop {-x\leq 9}} \right.\\\\\\\left \{ {{x\leq -4} \atop {x\geq -9}} \right.

                       /////////////////////////////////////////////////////

___________[ - 9]___________[ - 4]___________

\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

Ответ : x ∈ [ - 9  ;  - 4]

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