Questions about math in the third year of high school. Math problems in senior high school?

I also thought about this problem when I was in high school, first of all, the front multiple-choice questions should be done quickly, the method should be used flexibly, and it is not necessary to do the whole process, you can use a special method to bring in the method and a series of quick practices, and then fill in the blanks as much as possible, basically send points in front, there are two difficult points in the back, the first 2 questions of the big topic are very basic to ensure that they are all right, and the big questions behind should have the concept of step-by-step scoring, don't look at the type of question that you have not seen it and feel that it is difficult to have no confidence, the first few steps can still be scored, The next few steps are written to where it counts, and this is a score. In general, we should pay attention to the foundation, ensure that the basic score is not lost, the time should be allocated well, if the level of multiple-choice questions is good in 30 minutes, generally, about 40, fill-in-the-blank questions should have 30 minutes to do, and then there is about an hour, the first 2 big questions are 15 minutes, and the rest of the time try to do the rest of the questions!

The mobile phone party code word can't hurt it, ask for points!

1. First of all, ask a teacher or senior for a knowledge outline.

2. According to the outline, first understand the formula of each section, memorize it first, read the formula derivation again, and then do the corresponding exercises.

3. If you make a mistake, ask the teacher No matter how simple or difficult the question is.

4. Explain math problems to students if they have the ability to promote understanding.

5. Insist on doing mock questions, and after doing it, you must correct the answer and do it right on your back.

The key to mathematics is to practice more, in high school, there are so many types of questions, you have to learn a little live, you can't just die to do the problem, you have to learn to draw inferences from one example.

In fact, it is to do more questions, practice makes perfect.

The first question, choose , get a=n=2 (simpler, set the formula).

In the second question, the focus is on dividing the fraction into two parts (the second row is b n).

Let the equation of the parallel line y=x+m first.

The simultaneous elliptic equation is followed by 5x*2+2mx+m*2-4=0.

Since there is only one intersection point, we get the root number 5 of m=+-.

d = root number 10 2'And d = 3 root number 10 2'Take the shortest distance as the root number 10 2

Your answer is wrong.

It should be (9 4) multiplied by root number 3; Nine-fourths root number 3 method. 1. Find the ratio of two hexagons, and it can be seen that it is an equal proportional series.

Find the sum of its first n terms, and then n takes the limit n equal to infinity.

2. Reasoning. Let the hexagon area be a

1. Connect all adjacent diagonal lines of the hexagon (only one adjacent vertex corner) to obtain 12 triangles and a small hexagon in the outer circle.

2. Connect the diagonal lines of the small hexagon (two adjacent vertex angles) to get 6 equilateral triangles.

3. The area of these 18 triangles is equal.

4. That is, the ratio of the area of the large hexagon to the small hexagon is 3:1, that is, the area of the small hexagon is 1 3 a, and the area of all small hexagons is added to 1 2 a

From the problem, we find a=3 2 root number 3

a+1 2a=3 2 3 2 root number 3 = 9 4 root number 3

(1) The field of definition of f(x) is , let f(x)=

x-1 is a function of , then there is a real number a, b such that f(a-x) + f(a+x) = b

For any x constant that satisfies a-x d and a+x d, i.e.

a-xa+x-2=b, (b+2)(a2-x2)=2a, a=0, b=-2

There is a=0, b=-2, such that f(a-x)+f(a+x)=b is constant for any x≠ a is constant, f(x)=

x-1 is a function

2) Proof if g(a+x)+g(a-x)=

2a-x+t

2a+x+t

b is constant, then 2a+x+2a-x+2t=b(2a+x+t)(2a-x+t) is constant, that is, (1-bt)(2a+x+2a-x)=b(22a+t2)-2t is established, 1-bt=0, b(22a+t2)-2t=0, and t≠0, b=

t，a=log2|t|．

There are real numbers a, b such that g(x) is a function of

3) The image of the function h(x) is symmetrical with respect to the line x=m (m is constant), h(m-x)=h(m+x), and when m≠a, h(x+2m-2a)=h[m+(x+m-2a)].

h[m-(x+m-2a)]=h(2a-x)=h(a+(a-x)), and h(a+x)+h(a-x)=b, h(a+(a-x))=b-h[a-(a-x)]=b-h(x), h(x+2m-2a)=b-h(x),h(x)=b-h(x+2m-2a)=h(x+2m-2a)=h(x+4m-4a).

h(x) is a periodic function, and the period is 4m-4a

If m=a, then h(a-x)=h(a+x), and h(a-x)=b-h(a+x), h(a+x)=

b2, obviously h(x) is a periodic function

In summary, h(x) is a periodic function.

The sine theorem, one ratio comes out. Determine the angle c range from a angle range and angle b this would you wow? And then the comparison comes out, and you just need to find the maximum value of the sine interval.

If n-1 is the lower corner, then a1=a2=7

If n-1 is not the bottom mark, then there is no solution to this problem.

To sum up, I think the landlord made a typo?