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DaryaApart
@DaryaApart
July 2022
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17 БАЛЛОВ!!! Решите пожалуйста! [tex]12x \leq (x+3)^{2} [/tex]
[tex] x^{2} + 4x+5 [/tex] больше нуля
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oganesbagoyan
Verified answer
12x ≤ (x+3)² ;
12x ≤ x²+
6x +9 ;
x²+6x +9 - 12x ≥ 0 ;
x²-6x +9
≥ 0 ;
(x-3)² ≥ 0 ⇒ x ∈(-∞ ; ∞) или x∈ R.
* * * x любое вещественное число * * *
-------
x² +4x +5 > 0 ⇔(x+2)² +1 >0 тоже
для всеx вещественных чисел
.
x ∈ (-∞ ; ∞)
.
1 votes
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Verified answer
12x ≤ (x+3)² ;12x ≤ x²+6x +9 ;
x²+6x +9 - 12x ≥ 0 ;
x²-6x +9 ≥ 0 ;
(x-3)² ≥ 0 ⇒ x ∈(-∞ ; ∞) или x∈ R.
* * * x любое вещественное число * * *
-------
x² +4x +5 > 0 ⇔(x+2)² +1 >0 тоже для всеx вещественных чисел.
x ∈ (-∞ ; ∞) .