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2 TheXpert: "e^х - х^e >= 0 при х > 0" - это, очевидно, придется доказать.
e^x > x^e
x > e*ln(x), x > 0
x - e*ln(x) > 0, x > 0
f(x) > 0, x > 0, f(x) = x - e*ln(x).
f'(x) = 1 - e/x = 0, x = e
f(e) = 0, f(1) = 1, f(e^2) = e(e - 2) => f(e) -- min = 0, => f(x) >=0, x > 0.
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x = sqrt(x^x)
Zachod, Andrei!
x = sqrt(x^x)
Is this a non-trivial solution?
I have one, and it does seem to describe all the solutions. But I haven't proved that yet. And I don't remember how it works yet (I picked it up by accident, something is still in my memory).
Mathemat писал(а) >>
Is that a non-trivial solution?
Yes. And the expression itself would look like this: (x^x)^x = x^(x^2);
Hence x^x = x^2.
vegetate, одна великая страна так и не нашла в другой маленькой стране химическое оружие, хотя пол планеты "убедила" в том,
что оно там есть. Могу продолжить, привести и другие примеры глобального вранья, но не буду.
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Так, что там с расходом топлива, можно подробнее вашу точку зрения.
Я лично утверждаю, что данный аппарат не взлетел бы с Луны.
Table 3: Apollo lunar module weights.
Nominal weight of the lunar vehicle 14,710 kg.
Read on for yourself.
Да. А само выражение будет выглядеть так: (x^x)^x = x^(x^2);
Отсюда x^x = x^2.
Well if x^x = x^2, then x=2? Then it would be 2^4 = 4^2. Or did I misunderstand something again?
P.S. The parametrization there is non-trivial, you can't take it offhand.
9th grade, though:)))
Then do the math for yourself.
I have counted before and not only me, although I had different data. But, this data is "good" too.