A nonnegative integer with an absolute value less than 3 is___ ．

A nonnegative integer with an absolute value less than 3 is___ ．

The non negative integers with absolute value less than 3 are: 0, 1, 2, so the answer is: 0, 1, 2

The number of all integers whose absolute value is greater than 3 but less than 6 is

There are four and five in positive numbers
Negative numbers are - 4, - 5
So there are four

Y = loga (3a-1) is always positive, and the value range of a is obtained,
This is what I do. In the first case, when a > 1,
a> 1,3a-1 > 0, loga (3a-1) > 0
The second case is 0

a> 3a-1 > 1 a > 2 / 3 at 1:00
Results the intersection is a > 1
0

Given that the function y = loga (x) has Y > 1 in the interval [2, + ∞), then the value range of a is?

Obviously this is an increasing function
Because if it's a decreasing function
Then x > = 2
y1
So the minimum is loga (2)
Of course, he has to be greater than one
So log a (2) > 1 = log a (a)
2>a
So 1

If x belongs to [a + 2, a + 3], there is always | loga (x ^ 2-4ax + 3A ^ 2) | = 1, try to determine the value range of A
If x belongs to [a + 2, a + 3], there is always | loga (x ^ 2-4ax + 3A ^ 2) | = 1, try to determine the value range of A

If f (x) = loga [(x-3a) (x-a)] is meaningful and (x-3a) (x-a) > 0, then x > 3A or x3a or a + 3A > 0,
The symmetry axis of y = (x-3a) (x-a) is x = 2A

If x belongs to [a + 2, a + 3], there is always | loga (x ^ 2-4ax + 3A ^ 2) | = 1, try to determine the value range of A

If f (x) = loga [(x-3a) (x-a)] is meaningful and (x-3a) (x-a) > 0, then x > 3A or x3a or a + 3A > 0,
The symmetry axis of y = (x-3a) (x-a) is x = 2A

If the function f (x) = loga (x − 2aX) decreases monotonically on X ∈ (1,2), then the value range of a is______ ．

If f (x) = loga (x − 2aX) and monotonically decreases on X ∈ (1,2), then the outer function y = logau is a decreasing function, that is, 0 ＜ a ＜ 1 and X − 2aX ＞ 0 is constant on (1,2), that is, 1-2a ≥ 0

Let f (x) = loga (2a + 1) satisfy f (x) > 0 on the interval (- 1 / 2,0)
(1) Finding the value range of real number a
(2) Solving inequality f (x) > 1

If x ∈ (- 1 / 2,0), then 0

Given that f (x) = (2a-1) x + 1, x = 1 is a decreasing function on R, the value range of a is obtained

Y = loga (x), x > = 1 is a decreasing function, and 0

Given the set a = {0, 2}, B = {1, A2}, if a ∪ B = {0, 1, 2, 4}, then the real number a=______ ．

According to the meaning of the question, if a ∪ B = {0, 1, 2, 4}, then the set a or B must contain element 4, and if a = {0, 2}, B = {1, A2}, then A2 = 4, that is, a = ± 2; so the answer is ± 2