The known function f (x) = (1 + LNX) / X (1) Finding monotone interval of function f (x) (2) If the function f (x) is not a monotone function in the interval (T, t + 1 / 2) (T > 0), the value range of the real number T is obtained (3) If the inequality f (x) ≥ A / x + 1 holds when x ≥ 1, the value range of real number a is obtained

If the number a is 24, the least common multiple of a and B is 168, and the greatest common divisor is 4, then the number B is ()
The answer is 28
Prove how it is found
168*4/24=28
168=4*2*3*7
24=4*2*3
So the number B is 4 * 7 = 28
Given that the function f (x) = ax-x ^ 2-lnx is a decreasing function on (1, positive infinity), find the minimum value of G (x) = e ^ 2x-ae ^ X-1 on [ln1 / 3,0]
The value range of a is a > = 2, G (x) = e ^ 2x-ae ^ X-1, let e ^ x = t, and simplify it
G (x) = e ^ 2x-ae ^ X-1 is a decreasing function on [ln1 / 3,0]
The minimum value is taken
g(x)min=g(0)=-a,
Number a has 7 divisors, number B has 12 divisors. The least common multiple of number a and number B is 1728. What are the numbers a and B
Use segment division to find the divisor of 1728, and then form a and B respectively
The process is cumbersome, I take the manuscript
Because it's the least common multiple, plus the common divisor
Let f (x) = ax + 2, G (x) = a2x2-lnx + 2, where a ∈ R, x > 0. Is there a negative number a such that f (x) ≤ g (x) holds for all positive numbers x? If it exists, find out the value range of a; if not, explain the reason
Let H (x) = f (x) - G (x) = ax + lnx-a2x2 (x > 0) assume that there is a negative number a, such that f (x) ≤ g (x) holds for all positive numbers X. that is, when x > 0, the maximum value of H (x) is less than or equal to zero. H ′ (x) = a + 1x − 2a2x = − 2a2x2 + ax + 1x (x > 0) (9 points) let h ′ (x) = 0, we can get: x2 = − 12a, X1 = 1a (rounding) (11 points) when 0 < x ＜ 12a, H ′ (x) ＞ When x ＞ 12a, H ′ (x) ＜ 0, H (x) decreases, so h (x) has a maximum at x = − 12a, which is also the maximum. H (x) max = H (− 12a) ≤ 0. The solution is: a ≤− 12e − 34 (13 points), so the negative number a exists, and its value range is a ≤− 12e − 34 (14 points)
36 the least common multiple of all divisors is 36. (judge) why?
The least common multiple of the divisor of any positive integer is itself
Because the greatest divisor of this positive integer is itself, so their common multiple is not less than itself. Any number must be a multiple of its divisor. Therefore, the least common multiple of all divisors is 36
If f (x) = lnx-a2x2 + ax, f (x) is a decreasing function in the interval (1, + &), what is the value range of a?
Then you first seek the derivative. If the derivative is less than 0, it is a decreasing function. 1 / x + 2A ^ 2x + A. It's simple. Discuss it yourself
a> 1 or a
The maximum factor of a number is 36, and the number is______ All of its factors are______ The minimum multiple of this number is______ .
The maximum factor of a number is 36, this number is 36, all the factors of this number are 1, 2, 3, 4, 6, 9, 12, 18, 36; the minimum multiple of this number is 36. So the answer is: 36; 1, 2, 3, 4, 6, 9, 12, 18, 36; 36
If the zeros of the function f (x) = 3x-7 + ln & nbsp; X lie in the interval (n, N + 1) (n ∈ n), then n=______ .
Because f (1) = - 4 < 0, f (2) = ln & nbsp; 2-1 < 0, f (3) = 2 + ln & nbsp; 3 > 0, and f (x) is an increasing function on (0, + ∞), so in the interval (2,3), n = 2, so the answer is: 2
The prime factor of 18 is () and the four divisors of 18 are ()
Fill in the blanks
The prime factor of 18 is (2, 3, 3), and the proportion of the four divisors of 18 is (6 / 2 = 9 / 3)

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