When a is (), it is equal to 0

When a is (), it is equal to 0

When a is positive infinity, it is equal to 0

6 / (X-2) is a natural number. How many values satisfy the condition x

X-2 = 1 or 2 or 3 or 6
Then x = 3 or 4 or 5 or 8
therefore
6 / (X-2) is a natural number. There are four values satisfying the condition X

If (6 / x) - 2 is a natural number, then the value of the natural number x satisfying the condition is________ individual

6 / X-2 is a natural number
6/X-2≥0
6/X≥2
X≤3
So the value of the natural number x can be 3-2-1.0, but 0 cannot be used as a divisor, so there are only 1, 2 and 3

If x is a false fraction (x is a natural number), then there are () values of X that meet the conditions!

If x of 5 is a false fraction (x is a natural number), then there are (innumerable) qualified values of X, but if x is a natural number greater than 5

A is a natural number, 12 ／ a 1 / 2 = 12 × (). If a = 5, 12 ／ a 1 / 2=（
Please answer the questions.

A is a natural number, 12 ／ a 1 / 2 = 12 × (a). If a = 5, 12 ／ a 1 / 2 = (60)?

In 1 / 2 ＜ 5 / m ＜ 3 / 4, the natural number M represents has ()

When the molecule is completely reduced to 15, we get
15/30

There is a series of continuous natural numbers starting from 3: 3, 4, 5.., one of which is removed, and then the average of the remainder is 12.8, so what is the number removed?

The average of the rest numbers is 12.8, so the middle numbers in this series may be 12 and 13. 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22 and 23
A total of 21 numbers, the average is 13, the total is 21 * 13 = 273. After 17 is removed, the average of the rest is 12.8

In this paper, five consecutive natural numbers are arranged in descending order. It is known that one third of the sum of these five consecutive natural numbers is 18 more than the third number
To analyze and process

Let these five numbers be (A-2), (A-1), a, (a + 1), (a + 2)
So 1 / 3 * 5A = a + 18
2/3a=18
a=27
So these five numbers are 25, 26, 27, 28, 29

Let K be a natural number and Ka + B = 0, then | A / | B | - 1 | + | a | / B-2 | = 0
A.3 B.2 C.3+3/k D.2-2/k

It's a choice of three

There is a five digit number, which is divided by 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 13. The remainder of these 12 natural numbers are different. What is the five digit number?

Obviously: the remainder divided by 1 is 0, the remainder divided by 2 can only be 1, the remainder divided by 3 can only be 2, the remainder divided by 11 can only be 10, the remainder divided by 13 can only be 11 or 12, if the remainder divided by 13 is 12, then the original number n + 1 is a multiple of 123.101113, then n = [1,2,3,..., 11,13] * k - 1 = 13 * 11 * 9 * 8 * 7 * 5 K - 1