Question

log0,1 (x+4) + log0,1 (x-5) < -1
Треба терміново!

Answer 1

Ответ:

Log0,1(x+4) + Log0,1(x-5) = Log0,1([x+4]*([x-5]))< 1

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Log0,1([x+4]*([x-5])) < 1

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Log0,1([x+4]*([x-5])) = [x+4] * [x-5] < 1 \

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([x+4] * [x-5] < 1)