Math problems are difficult to solve by experts

2 Yes"square"~

Solution: By known conditions:

x+y+z=2

x^2+y^2+z^2=3

So xy+yz+zx=(1 2)[(x+y+z) 2-(x 2+y 2+z 2)]=1 2

And because of the first term on the left.

1/(xy+z-1)=1/[xy+(2-x-y)-1]=1/[(x-1)(y-1)]

In the same way, 1 (yz+x-1)=1 [(y-1)(z-1)]1 (zx+y-1)=1 [(z-1)(x-1)]The addition of the three formulas (in this case the general division is very simple) obtains:

3-x-y-z)/[(1-x)(1-y)(1-z)]1/[(1-x)(1-y)(1-z)]

The conditions of the questions are incorrect and conflicting.

x-square, y-square, z-square=3

by arithmetic-geometric mean inequalities.

3=x-square, y-square, z-square"3 cubic root number under [(xyz) 2], so xyz "1

The condition is xyz=1 again

So the inequality equal sign holds.

Launch x=y=z=1

So x+y+z=3 instead of x+y+z=2, the condition is wrong.

1/(xy+z-1)+1/(yz+x-1)+1/(zx+y-1)

The answer on the ground floor is correct

I've done this before

The 2nd and 3rd floors don't talk nonsense

The answer on the first floor is very correct

I think that's right.

Suppose the number of hundreds, 100a+10b+c, where a, b, and c are integers from 0 to 9 and a is not equal to 0

Then the problem translates to a known a+c-b=11d, where d is an integer.

Prove that 100a+10b+c is a multiple of 11.

It is not difficult to derive -101 from the hypothesis, when d=0, a+c=b; 100a+10b+c=99a+11b=11*(9a+b), 9a+b is an integer from the assumption, and it is true!

2. When d=1, a+c=b+11;100a+10b+c=99a+11b+11=11*(9a+b+1), 9a+b+1 is an integer from the assumption, and it is true!

In summary, the conclusion is valid!

Question: You and your brother received a red envelope during the Chinese New Year, the red envelope can only be a certain value of 20, 40, 80, 160, 320, 640, and your two money is different, and your two money is connected together (for example, if you get 80 and your brother is 40 or 160), assuming that you and your brother are smart enough, you can't see each other's money, but you can talk, and you can exchange red envelopes after reaching an agreement, the question is that now you have 80 in the red envelope, will you finally exchange it with your brother? Why?

A: There will be no exchange.

2) Let's say you get 320, aIf you ask your brother first, and he answers no, then he is 640, and if he is not sure that it is 160, you will not continue to ask, and you will definitely not exchange bIf he asks you first, he must be 160, and you will be sure that you will not exchange with him at once.

3) Let's say you get 160, aAsk your brother first, if he is 320 then the same (2b.).He won't exchange, if he's 80, he's not sure, and you're right away sure he's 80, and he's not going to exchange b

Brother asks you first, you are not sure, if he is 320 then you are sure that you are 160 and cannot be exchanged, if he is 80 then you will ask him again, he is still not sure that you can be sure that he is 80, and you will not exchange with him.

4) Now you're 80, aYou ask first, and he's not sure, and then he asks you if he's not sure, if he's 160, according to (3b.).He's sure you're 80, and if he's 40, he'll ask you again if he's going to change it, and you're sure he's 40 and can't exchange b

He first asked if he was 160, then the same (3a.)If he is 40, he asks first, you are not sure, you ask again, he is not sure, you are sure that he is 40, otherwise he is 160 as early as the first time you are not sure, you are sure that you are 80 and do not want to exchange.

So in the end, it is impossible to exchange red envelopes.

It's a bit messy to write, this belongs to logical mathematics, or a problem in game theory, I don't know if it meets your requirements.

1. The last three digits of the number 1978 n are equal to 1978 m, try to find the positive integers m and n, so that m+n takes the minimum value, where n > m 1

Find one get one free: 2, let 0 a b c d e, and a+b+c+d+e=1, and verify ad+dc+bc+be+ea 1 5 (this question is super simple).

Sure enough, it was a difficult ......Untitled is better than untitled.

Proof: divisible space and compact space do not contain each other.

The topic is not rigorous:

1) The phrase "pine and birch multiple hectares" does not say how hectares are allocated to "pine and birch"?

Or (2) the phrase "poplars are still more hectares than birch" does not say that there is more area planted than the original birch trees or is it more than the new planted area?

Pine accounts for all = 1 (1+7) = 1 8

Birch accounts for all = 3 (3 + 7) = 3 10

Altogether = hectares.

You're so stupid, I'm in fifth grade right now. This question is a piece of cake for me. You can't just rely on others to do your own thing!!

c。Solution: x1 <0, x2-3x1<0 get x1+x2<4x1<0, x1x2>3(x1) 2>0

Because the equation x 2 (1 m) x n 2 0 has two different solutions, which is derived from Veda's theorem: x1 x2 m 1 0

x1x2＝n－2＞0

So m 1, n 2

c。Because x1<0, x2<0It can be solved by Vedder's theorem.