Math proof questions. It s super hard.

pe=xpf=y

ef=zpe*eb=2zz

pf*fc=2zz

eb=2zz/x

fc=2zz/y

For the sake of convenience, let's assume z 1, otherwise we can zoom in or out of the whole graph in proportion.

pe=xpf=y

ef=1pe*eb=2

pf*fc=2

eb=2/x

fc=2/y

Cosine theorem. ab2 = 1 + (2 x) (2 x) - 2 (2 x) cos angle AEB

yy=1+xx-2xcos angle PEF=1+xx-2xcos angle AEB

ab^2=1+(2/x)(2/x)+(2/xx)[yy-1-xx]

1+4/xx-2/xx+2yy/xx-2

2/xx+2yy/xx-1

Similarly. cd^2

2/yy+2xx/yy-1

1=xx+yy-2xy*cos angle epf

pb=pe+eb=x+2/x

pc=y+2/y

BC 2 = (x+2 x) (x+2 x) + (y+2 y) (y+2 y) -2 (y+2 y) (x+2 x) * cos angle bpc

Angular BPC = Angular EPF

So. bc^2=(x+2/x)(x+2/x)+(y+2/y)(y+2/y)+(1+2/yy)(1+2/xx)(1-xx-yy)

ab·cd=ef·bc

Evidence. ab·cd ab·cd=ef·bc ef·bc is fine.

ab^2=2/xx+2yy/xx-1

cd^2=2/yy+2xx/yy-1

bc^2=(x+2/x)(x+2/x)+(y+2/y)(y+2/y)+(1+2/yy)(1+2/xx)(1-xx-yy)

ef^2=1

ab^2*cd^2=bc^2*ef^2

So. ab*cd=bc*ef

This problem is simple, convert to a proportional calculation. Now I'm off work, make a mark, and solve this question tomorrow It's not complete! Or the wrong topic.

First of all, the following questions are obtained from the question:

1. The positions of the four points a, b, c, and d are arbitrary;

2. P is not necessarily on a circle.

3、ae=ef=fd

So suppose that the quadrilateral ABCD is a square (not contrary to the requirements of the question) and the side length: 3 can be obtained: ae=ef=fd=1

Then: ab*cd=3*3=9, ef*bc=1*3=3ab·cd≠ef·bc

The parity of the function can be judged according to the following equation.

f(-x) = f(x), and f(x) is an even function.

f(-x)=-f(x), f(x) is an odd function, as previously known as the slag.

The proof process is as follows:

When solving the inequality, we first need to find the relationship of the inequality, and then look at the limited threshold range of the unknown to condition the relationship.

<> look at a source of hail containing this treasure old stool.

Define the domain as closed r

f（x）+f（-x）

ln（x+/x^2+1）+ln（-x+/x^2+1）ln（x^2+1-x^2）ln1

Then f(-x) = -f(x) triumphant annihilation.

The function is an odd function, which is proven.

According to the addition formula of probability, there are p(ab)+p(ac)-p(abc)=p(a(buc))<=1

That is, p(ab)-p(abc)<=1-p(ac) and p(abc)<=p(bc), so p(ab)-p(bc)<=p(ab)-p(abc)<=1-p(ac).

In the same way, it can be proved that p(bc)-p(ab)<=1-p(ac) has |p(ab)-p(bc)|<=1-p(ac)

You can try to do this problem using mathematical induction (I'm just guessing).

Let this positive integer be a, then a=1,2,3....

When a = 1, 3a + 1 = 4, 4 2 = 2, 2 2 = 1, (and then there is a 4, 2, 1 cycle).

When a=2, (ditto).

When a=3, (e.g. solution).

When a=1, (same as a=1).

It can be judged that a is a multiple of 4 as long as the final calculation is based on the above algorithm.

The following is proved by mathematical induction:

When a = 1, 3a + 1 = 4, 4 2 = 2, 2 2 = 1, (and then there is a 4, 2, 1 cycle).

Assuming that this is also true when a=n, then 3n+1=m (let m be an integer multiple of 4).

Then, when a=n+1, 3(n+1)=3n+1+3=m+3

m+3 must be odd, 3 (m+3)+1=3m+10

For 3m, which is an integer multiple of 4, the above conclusion must be made.

For 10, 10 2 = 5, 5 3 + 1 = 16, 16 2 = 8, 8 2 = 4, 4 2 = 2, 2 2 = 1, 1 3 + 1 = 4, and then this cycle.

Proof of proof. This is the famous "Koalaz conjecture", in fact, it has not been mathematically proven to be completely true, (hehe, I also made this nonsense, I think it's about the same!) ）

However, it is contemptible that as long as the prime number can be proved, it should be possible. However, it seems that this project is relatively vast!

3a 1 2a a 1, if it is an odd number, then the result must be an even number, divided by 2, it is a (a 1) 2, at this time (a 1) 2 has two possibilities of odd and even numbers: if (a 1) 2 is an odd number, also a (a 1) 2 is an even number, perform the next step, divide by 2; If (a 1) 2 is an even number, then a (a 1) 2 is an odd number, and then 3 1 gets 4a 2 (a 1) 2, which is an even number, and then divides by two 2a 1 (a 1) 4, the latter is simple, do your own counting.

This is. Kouraz guessed, it's too hard. ~！

In fact, whether it is proof questions, or answer questions, if you want to do a good job, the most important thing is to find ideas, many students often get overwhelmed when they encounter proof questions, and begin to let their thinking diverge infinitely, and finally make themselves confused. Therefore, for a proof problem, we must first dig out the known conditions, find some hidden conditions, and then determine the method according to the known conditions, how to determine the method? This needs to be continuously accumulated, and some will be summarized in the textbooks and teachers' explanations, such as the practice of common auxiliary lines in junior high school geometry, the methods of counterproof and mathematical induction in high school, and the basic methods of various limits, calculus, and matrix problems in college.

Only in this way, when you see a question, it is possible to know it in your heart.

If you don't want to practice, don't want to summarize, don't even want to listen to lectures, and don't want to learn, then there are also methods: 1. Turn yourself into a genius, but it seems that geniuses like to study; 2. When encountering a question that will not be introduced, all the known conditions and the implicit conditions of the introduction are listed, a curly brace, and the answer is directly introduced, but this method is only useful for a small number of questions and a small number of teachers, and most of the time, the reviewer will send a big cross.

My words are generally from simple to difficult, and you can do a good job only if you have a solid foundation to prove the questions, and looking at some questions that are easy to make mistakes will improve your level

Two lines of thought, one against the law, and one on the positive evidence.

The drawings are not standard, and they can be analyzed according to the drawing

These 3 triangles can be regarded as triangles with cd as the bottom edge, respectively through the points a, b, m to achieve the height of cd, through similar triangles can prove ah + mh = bh (every two triangles are similar, and the title has ao + m0 = bo), the triangle area is 1 2 bottom edge * high, the bottom edge is the same, the height is brought into the above formula, there can be.

s△dmc=s△dbc-s△dac。

There is one less condition for this question: an is the equal difference Tongzhou Xianglie.

Use an=a1 + n-1 d

sn=na1+n(n-1) 2 d is brought into the left and right of the local beat of the required formula, which will be found.

Left = right trace thick edge = 2a1 + m + n-1) d.

Why is there an S on the left and no S on the right?

What, is an arbitrary sequence?? Let respect?? Proportional, Equal or Other.

Note that asking questions is also about the level of frankness and prudence.

Not true! The proof is as follows: a+ b+ c=180° from abc gives a+ 1= 3 by the formula for the complementary angle of the triangle

The sum of the internal angles of the trapezoid is 360° b+ c+ 2+ 4=360°, i.e. b+ c+ 2+(180- 3)=360°b+ c+ 2+(180- a- 1)=360°180° -a+ 2+(180- a- 1)=360°: 2- 1=2 a See if you can understand?

No, 2- 1=2 a, can be proved by the sum of the outer angles.

You try it with the sum and difference product, and the sum and difference formula.