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masha20009
@masha20009
July 2022
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икс квадрате минус 64 меньше 0
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X²-64<0 ⇔ (x-8)(x+8)<0
+ - +
-------------(-8)//////////////////(8)------------
x∈(-8;8)
1 votes
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TARTILLO
Verified answer
Х(2)-64 < 0 при х < 8, т.к. при х=8 64-64 даст 0, что будет неверно в исходном неравенстве.
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+ - +
-------------(-8)//////////////////(8)------------
x∈(-8;8)
Verified answer
Х(2)-64 < 0 при х < 8, т.к. при х=8 64-64 даст 0, что будет неверно в исходном неравенстве.