In rational numbers, the opposite number and reciprocal number are equal to themselves: a.0 B.1 C. - 1 D. does not exist, which one to choose?

In rational numbers, the opposite number and reciprocal number are equal to themselves: a.0 B.1 C. - 1 D. does not exist, which one to choose?

D. It doesn't exist

If a + B = 0, then a and B must be ()
A. They are all equal to 0b. One positive and one negative C. they are opposite numbers D. a > B

∵ a + B = 0, ∵ a, B are opposite to each other

A negative number minus its opposite number, and then divided by the absolute value of the negative number, the quotient is______ ．

When a ＜ 0, [a - (- a)] / | - a | = [a + a] / (- a) = 2A / (- a) = - 2. So the quotient is - 2. So the answer is: - 2

If the absolute value of one number is 8, the absolute value of the other number is 4, and the product of the two numbers is negative, find the quotient of the large number divided by the decimal

Let | a | = 8, | B | = 4, then a = ± 8, B = ± 4, ∵ AB < 0, ∵ when a = 8, B = - 4, ∵ 8 − 4 = - 2, when a = - 8, B = 4, ∵ 4 − 8 = - 12

There is a fraction. If the numerator is reduced by 1, then the fraction becomes 13. If the denominator is reduced by 1, then the fraction becomes 12, then the fraction is 13______ ．

According to the meaning of the question, the denominator is three times less than the numerator, and the denominator is two times more than the numerator; the numerator of this fraction is: (1 + 3) / (3-2) = 4, the denominator is: 3 × 4-3 = 9, and the fraction is 49

There were 570 students in a school last year and 610 in this year. This year, the number of boys increased by 1 / 6 of last year, while that of girls decreased by 1 / 5. How many more boys than girls last year
This is a small six to fill in the blanks, prawns have no direct way to calculate the number of boys more than girls

If there were x girls, there were 570 - x boys
4/5x+(570-x)*6/7=610
5/4x+570*6/7-6/7x=610
5/4x+665-6/7x=610
665-(6/7x-5/4x)=610
665-(30/35x-30/11x)=610
665-30/11x=610
665-610=30/11x
55=33/11x
x=55*11/30
x=150
570-x=570-150=420
420-150=270
A: 270 more

A pear has 350 kg, two fifths more than an apple. How many kg does an apple have

Apple x + 2 / 5 (two fifths) x = 350 x = 250

There are 240 students in the lower grade of the experimental primary school, and the number of students in the middle grade is equivalent to three-quarters of the lower grade and six seventh of the higher grade?

240 times 3 / 4 divided by 6 / 7 is 210 people

In permutation and combination problems, when to divide by factorial and when not to divide
QT, such as: 6 books divided into a, B and C
1. Two copies per person
2. A 1, B 2, C 3
3. One for one, two for one, three for one
It mainly talks about when to divide factorial when to combine and when not to divide

*It's a multiple sign
1: First of all, C6 (2) 2 * C4 (2) * C2 (2). The calculated result is equivalent to the stipulation that AB is in the first pile, CD is in the second pile, EF is in the third pile. But in the question, there are two books per capita, which means that you don't want to sort, but the above formula is obviously in the order, so to break the order, you have to divide it by 3!. if you don't understand, you can think about it backwards, assuming that you have divided the average three piles, But if you want to divide it among three people, you have to multiply it by 3! 1, so you don't want to sort it, so divide it by 3!
2: From the title, we know that the order has been arranged, which stipulates that a person should take one copy, B two copies and C three copies. Therefore, C6 (1) * C5 (2) * C3 (3). This formula is equivalent to stipulating that the first pile of one copy, a can take it, the second pile of two copies, B can take it, and the third pile of three copies, C can take it
3: According to the formula in 2, there are three piles. The meaning of the problem is that there is no fixed owner for the three piles of books. A can take any pile. There are three situations. After a takes it, B has two situations and C has one situation. Therefore, formula 2 has to be multiplied by 3 * 2 * 1!
So, to sum up, the average divided by the factorial of the number of shares, the uneven multiplied by the factorial of the number of shares, the provisions of a person to take a pile, do not deal with
High school students, right? I've been forced to be a junior in high school, and my grades are OK. It's not hard to understand what I said to my classmates. I hope it's useful for you. Come on, never give up until the last moment!
It's useful for others to see it
Shit! 10 years? I'm late

The question of permutation and combination about the problem of equal distribution
There are two topics
1) There are four players in the bridge competition. How many points are there when each player gets 13 cards?
2) There are 12 people, divided into two groups, 6 people in each group, how many points?
3) Ten children are divided into two groups, five in each group. How many points are there in total?
What's the difference between these three problems? Which one needs to divide by the factorial of the group, which one doesn't? Why?

The main idea of the three problems can be the same. Use factorial to represent the permutation of all numbers, and then divide by the permutation in each group. The difference in details is: whether the groups are differentiated
1.52! / (13!) ^ 4 / / four players are distinguished. That is to say, it is different to exchange the cards of two players
3.10! / (5!) ^ 2 / 2 = C (10,5) / 2 / / there is no distinction between the two teams. That is to say, all the players of team a and team B are exchanged, and they are regarded as the same group
2. The question is indefinite
12! / (6!) ^ 2 = C (12,6) / / if the two groups are distinguished
12! / (6!) ^ 2 / 2 = C (12,6) / 2 / / if the two groups are regarded as indistinguishable
The last division of 2 and 3 is because the two groups are indistinguishable