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znikolenko
@znikolenko
July 2022
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укажите область определения функции y=корень из (3-x)(x+7)
решите пожалуйста
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Answers & Comments
джек005
Y=√((3-x)(x+7))
y²=0
3-x=0, x+7=0
x=3, x=-7
[-7;3]
1 votes
Thanks 1
nKrynka
Решение
y =√
(3-x)(x+7)
(3-x)(x+7)
≥ 0
(x - 3)*(x + 7) ≤ 0
x - 3 = 0
x₁ = 3
x + 7 = 0
x₂ = - 7
D(y) = [- 7; 3]
2 votes
Thanks 1
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Answers & Comments
y²=0
3-x=0, x+7=0
x=3, x=-7
[-7;3]
y =√(3-x)(x+7)
(3-x)(x+7) ≥ 0
(x - 3)*(x + 7) ≤ 0
x - 3 = 0
x₁ = 3
x + 7 = 0
x₂ = - 7
D(y) = [- 7; 3]