Sveta11111111Love
17.05.2022 13:54

Представьте выражение в ви де многочлена:​


Представьте выражение в ви де многочлена:​

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Ответ:
23LOLKEK
16.06.2022 01:58

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Must Waste More Time...

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Ответ:
ПАПА1111111111
07.02.2021 18:01

|x + 2|(x² – a²) > 0

1) a ≤ –2: x ∈ (–∞; a) ∪ (–a; +∞)

2) –2 < a < 0: x ∈ (–∞; a) ∪ (–a; +∞) \ {–2}

3) a = 0: x ∈ (–∞; +∞) \ {–2; 0}

4) 0 < a < 2: x ∈ (–∞; –a) ∪ (a; +∞) \ {–2}

5) a ≥ 2: x ∈ (–∞; –a) ∪ (a; +∞)

Объяснение:

Выражение |x + 2|(x² – a²) -- может менять знак только в точках, являющихся корнями уравнения |x + 2|(x² – a²) = 0, то есть корни делят числовую прямую на интервалы, в пределах которых знак сохраняется.

Для решения неравенства |x + 2|(x² – a²) > 0 необходимо нанести корни на числовую прямую и пометить те интервалы, на которых выражение |x + 2|(x² – a²) является положительным. Сами корни не будут входить в ответ, поскольку неравенство строгое.

Корнями являются значения x₁ = –2, x₂ = –a, x₃ = a. Существует несколько возможных вариантов расположения этих корней на числовой прямой, поэтому необходимо рассмотреть их все по отдельности (см. рисунок).


Номер 4.1 уравнение с параметром (если не трудно,объясните как решать подобные уравнения с параметро
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