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mokek
@mokek
October 2021
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решите пожалуйста) с решением
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sukonnikoffma
(x^2-4x)^2-16>=0
(x^2-4x-4)(x^2-4x+4)>=0
(x^2-4x+4-8)(x-2)^2>=0
((x-2)^2-8)(x-2)^2>=0
(x-2-2sqrt2)(x-2+2sqrt2)(x-2)^2>=0
ответ:
(-беск;2-2sqrt2] u {2} u [2+2sqrt2; +besq)
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Answers & Comments
(x^2-4x-4)(x^2-4x+4)>=0
(x^2-4x+4-8)(x-2)^2>=0
((x-2)^2-8)(x-2)^2>=0
(x-2-2sqrt2)(x-2+2sqrt2)(x-2)^2>=0
ответ:
(-беск;2-2sqrt2] u {2} u [2+2sqrt2; +besq)