• Предмет: Математика
  • Автор: ilona79792004
  • Вопрос задан 1 год назад

|х+1|+|х+2|+|х+3|=6
даю 20 балов ​


MrOligarch: i have a solution but i can't send why?
CodedEmerald: you can do it now, try
CodedEmerald: previous answer was incomplete

Ответы

Ответ дал: MrOligarch
1

Пошаговое объяснение:

follow for more solution Mr.Oligarch

Приложения:

MrOligarch: if my answer is helpful to you please vote as brainlienst
Ответ дал: CodedEmerald
3

Ответ:

x₁ = -4

x₂ = 0

Пошаговое объяснение:

|х + 1| + |х + 2| + |х + 3| = 6

Обнулим модули:

x + 1 = 0

x = -1

================

x + 2 = 0

x = -2

================

x + 3 = 0

x = -3

Находим множество доступных вариантов для x:

x ∈ (-∞; -3] ∪ (-3; -2] ∪ (-2; -1] ∪ (-1; +∞)

================================

\left[\begin{array}{ccc}\left \{ {{x \in (-\infty; -3]} \atop {-x-1-x-2-x-3 = 6}} \right.\\\left \{ {{x\in(-3; -2]} \atop {-x-1-x-2+x+3=6}} \right.\left\\\left \{ {{x\in(-2;-1]} \atop {-x-1+x+2+x+3=6}} \right.\\\left \{ {{x\in(-1; +\infty)} \atop {x+1+x+2+x+3=6}} \right. \end{array}

\left[\begin{array}{ccc}\left \{ {{x \in (-\infty; -3]} \atop {-3x-6=6}} \right.\\\left \{ {{x\in(-3; -2]} \atop {-x=6}} \right.\left\\\left \{ {{x\in(-2;-1]} \atop {x+4=6}} \right.\\\left \{ {{x\in(-1; +\infty)} \atop {3x+6=6}} \right. \end{array}

\left[\begin{array}{ccc}\left \{ {{x \in (-\infty; -3]} \atop {x=-4}} \right.\\\left \{ {{x\in(-3; -2]} \atop {x=-6}} \right.\left\\\left \{ {{x\in(-2;-1]} \atop {x=2}} \right.\\\left \{ {{x\in(-1; +\infty)} \atop {x=0}} \right. \end{array}

\left[\begin{array}{ccc}x=-4\\x\in\varnothing\\x\in\varnothing\\x=0\end{array}

x ∈ {-4; 0}

x₁ = -4

x₂ = 0

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