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mobbydyk
@mobbydyk
August 2021
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Под корнем x+7 больше x+1
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infor5
ОДЗ x+7≥0 ⇒x≥ - 7
возводим в квадрат обе части
x+7>x^2+2x+1
x^2+x-6<0
(x-2)(x+3)<0
......+..........-3....-......2..........+...........x
Ответ x∈( -3 ; 2)
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Answers & Comments
возводим в квадрат обе части
x+7>x^2+2x+1
x^2+x-6<0
(x-2)(x+3)<0
......+..........-3....-......2..........+...........x
Ответ x∈( -3 ; 2)