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ivannikolaevich1
@ivannikolaevich1
July 2022
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Помогите с алгеброй, умоляю!
Найдите область определения функции y=12x-3/3-6x^2
На фотке 4 номер.
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NSema
3-6x^2 не = 0
-6х^2 не = -3
2х^2 не = 1
х^2 не = 1/2
х не = 1/√2 и -1/√2
х= (- безконечность; -1/√2) и (-1/√2; 1/√2) и (1/√2; + безконечность)
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Answers & Comments
-6х^2 не = -3
2х^2 не = 1
х^2 не = 1/2
х не = 1/√2 и -1/√2
х= (- безконечность; -1/√2) и (-1/√2; 1/√2) и (1/√2; + безконечность)