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derekmetyus
@derekmetyus
October 2021
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область определения функции:
(x^2-5x+6)^(1/2)+(x-7)^(1/3)
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kirichekov
Verified answer
x²-5x+6≥0
++++[2]--------[3
]++++++++++>x
ответ:
x∈(-∞;2]∪[3;∞)
2 votes
Thanks 0
derekmetyus
а разве на кубический корень существуют ограничения?
kirichekov
спасибо, исправлю.
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Answers & Comments
Verified answer
x²-5x+6≥0
++++[2]--------[3]++++++++++>x
ответ: x∈(-∞;2]∪[3;∞)