The last question in the mathematics paper of the junior high school graduation exam

The method is a bit cumbersome, but it's also 3 8.

Let the coordinates of a move be (a -a+3).

Calculate the area of the overlapping part s=-a (squared) 2+3a 2-3 4 When the coordinates of a move are (3 2 3 8), the maximum area is: 3 8 The figure is not very clear, sorry, the computer took it).

The topic should not be written completely, just give it so far, I personally think that the second question should be when point A coincides with point C, the area of the triangle AOD and the triangle OCB coincides is the largest, that is, the area of the triangle OCD is equal to the area of the triangle OCB - the area of the triangle CDB = 1 * 2 * 1 2 - 1 * 1 * 1 2 = 1 2.

Who are you fooling??? The first question has three point coordinates, and it is simple to find the analytic formula.

Second question: Isn't that the area of that triangle??? How easy is 2*1*???

Maybe it's like the brother below said that the topic is not complete.

Isn't hg y=-x+3?

In this case, the maximum value is indeed 3|8, at this time OA is Y=2X-3 2, you are right!!

Give it points!

The area of the triangle OCD is equal to the area of the triangle OCB - the area of the triangle CDB = 1*2*1 2-1*1*1 2=1 2.

Did you make a mistake in your question, how did I calculate that when a and c coincide with the largest area, you can see that it is the largest when you look at the picture.

Uh, may I ask you to take the test so soon? I won't take it until the day after tomorrow. Looking at the computer, I can't figure it out. I had to look at the exam paper to be inspired.

There are answers and steps on the Internet?

I just came from the third year of junior high school, and the last question is generally a maximum of 10 points, and there are three questions in total. The first group of good Lu asked you to be sure that your socks were wide, and if you couldn't do the last two questions, you would give up and go back to check. In this way, you trade 7 points for at least 100 points of accuracy.

1, 2nd floor wrong!

Idea: Make point B about the point M symmetry of AC, connect MD, and cross AC to P, at this time, the point P makes the crack value of L the smallest (according to the theorem of the perpendicular bisector and the shortest line between two points).

I'm a little wrong upstairs should be: pass the point M as the X-axis perpendicular intersection x in N, let the BM source leak AC in Q, it is easy to know that Q is the midpoint of AC, so MBN=30°, so mn=1 2BM=AD=2 3, by the Pythagorean theorem to find bn=6, so cn=2, and by the Pythagorean theorem can find the minimum value of L, that is, the length of DM, the coordinates are easy to find

I'll copy the answer downstairs to you for your reference.

The two vertices are on ab and ac respectively. Q: What is the side length of the machined square part? Xiaoying solved the answer as 48mm, and asked the following questions.

1) If the part to be machined in the original problem is a rectangle, and this rectangle is composed of two squares placed side by side, find the side lengths on both sides of the rectangle. (2) If the part to be processed in the original problem is only a rectangle, in this way, the length of the two sides of the rectangle can not be determined, but the area of the rectangle has a maximum value, and when the maximum value is reached, the length of the two sides of the rectangle.

Solution: (1) C diesel engine: (10-x-y) set.

2) Because 32 acres of farmland are irrigated, a A diesel engine needs to be equipped with 4 pumps, so A works for one hour, irrigates 4x acres, and a B diesel engine needs to be equipped with 3 pumps, so B irrigates 3y acres in one hour, and a C diesel engine needs to be equipped with two pumps, so C irrigates 2 (10-x-y) mu in one hour, then there is 3x+4y+2(10-x-y)=32 solution: y=12-2x

2.Total cost: w = 130x + 120y + 100 (10-x-y) = 30x + 20y + 1000 = 30x + 20 (12-2x) + 1000 = 1240-10x. Because as long as there is one of each, then there is x greater than or equal to 1, y greater than or equal to 1, and 10-x-y greater than or equal to 1, and the solution results that x is greater than or equal to 3 and less than or equal to, so when the minimum cost is x=3, the cost is 1210

1，（2）

Solution, derived from the meaning of the question:

At the same time, it works for one hour, irrigates 32 acres, and each pumping machine can pump water to irrigate 1 mu of farmland per hour, a total of 32 pumping machines.

4x+3y+2(10-x-y)=32

x,y,(10-x-y) are all greater than 1

So, to simplify: y=12-2x where (3 x 5, and x is an integer)2, y=12-2x

Both: To irrigate according to the requirements, there are a variety of combinations of y and x, as long as y = 12-2x is satisfied, A is 130 yuan per hour, which is more than B.

Try to arrange as little as possible.

x minimum is 3

3 sets of SO, 6 sets of B, 1 set of C.

Don't scold me if you do something wrong.!!

If there are x standard rooms, y rooms, double rooms, z rooms, single rooms, then according to the "current tour group stays in 20 rooms", we get x+y+z=20 At the same time, we can list the system of equations for the total number of people: 3x+2y+z=50 Let's first sort out a group of binary inequalities, substituting with 2x+y=30 At the same time, 3 - get y+2z=10 Total ** should be: p=60 3x+100 2y+200 z simplification to get p=180x+200y+200z we want to find the minimum value of 180x+200y+200z, and the equation should satisfy the substitution of will into this equation, get 20y+20z+3600=p, and then substitute into , get 10y+3700=p, at this time we want the y value to be the smallest, so let's take a look at the minimum value of y, note the :

2x+y=30, so the minimum value of y can be 0 (when x goes to the extreme value of 15), then the lowest is 0 y+3700=3700 yuan.

Suppose that the triple room, the double room and the single room are divided into x,y,zx+y+z=203x+2y+z=50 respectively, so that k=3 60x+2 100y+200z is the smallest x,y,z. Note that k = 200 (x + y + z) - 20x, so just ask for the maximum x. Simplification has x+y+z=20, 2x+y=30 by the second formula, x can only be 15 at most, and y needs to be 0 at this time, then z is 5, and this situation exists, so this is the mode of the lowest cost.

x=15,y=0,z=5.Bringing in k can be used to calculate the minimum**.

Set the number of single rooms for x, the number of double rooms for y, the number of triple rooms for z, the total accommodation cost of this tour is b (yuan), from the question can be known: x+y+z=20 (1) x+2y+3z=50 (2) 200x+200y+180z=b (3) multiplied by (1) multiplied by 200 into (3) formula simplification can be obtained: 4000-20z=b (4) subtract (1) from (2) to obtain:

y+2z=30 (5) Because y is greater than or equal to 0 and x is greater than or equal to 0, we can get from equation (5): 10=< z<=15 (6) Substituting (4) into equation (6) can give the inequality with the smallest value of b 4000-b<=300, that is: b>=3700 Therefore, when b takes the minimum value of 3700 yuan, at this time z=15, x=5, y=0 Answer:

5 people in 5 single rooms and 45 people in 15 triple rooms make the travel cost the lowest.

1. Calculation and proof of line segments and angles.

The answers to the questions in the high school entrance examination are generally divided into two to three parts. The first part is basically a series of simple or intermediate questions, which are designed to examine the basics. The second part is often the middle problem of starting to pull points.

The significance of easily mastering these questions is not only to get scores, but more importantly, to affect the morale and morale of the army throughout the process of doing the questions.

2. Quadratic equations and functions.

Among these problems, the dynamic geometry problem is the most difficult. The difficulty of geometric problems lies in imagination and construction, and sometimes an auxiliary line is not thought of, and the whole problem is stuck. Compared with the geometry comprehensive questions, the algebra comprehensive questions do not require too many ingenious methods, but they have relatively high requirements for the candidates' calculation ability and algebra skills.

In the mathematics of the high school entrance examination, algebra problems are often based on unary quadratic equations and quadratic functions, and a variety of other knowledge points are assisted. In the problem of unary quadratic equations and quadratic functions, the pure unary quadratic equation solution method is usually investigated in the form of simple solution. However, in the later difficult questions, it is usually combined with knowledge points such as discriminant roots, integer roots and parabolas.

3. Cross-synthesis of multiple functions.

The functions involved in junior high school mathematics are primary functions, inverse proportional functions, and quadratic functions. This kind of question itself is not too difficult, and rarely appears as a finale question, and is generally used as a mid-level question to test the candidate's mastery of the primary function and the inverse proportional function. Therefore, in the face of this kind of problem in the high school entrance examination, we must avoid losing points.

4-column equation (group) solution problem.

In the high school entrance examination, there is a type of question that is not difficult to say, not difficult and difficult, sometimes there are ideas in three or two, and sometimes there are no ideas after thinking and meditating for a long time, which is the application problem of solving column equations or systems of equations. Equations can be said to be the most important part of junior high school mathematics, so they are also a compulsory content in the high school entrance examination. Judging from the high school entrance examination in recent years, there are more exams combined with current affairs, so candidates also need to have some life experience.

In the actual exam, this kind of question almost always gets a full score or no score, but there are only a few types of questions, so candidates only need to practice more and master each question category and summarize some formulas, and they can deal with it calmly.

5. Dynamic geometry and function problems.

On the whole, there are probably two emphases of algebraic comprehensive problems, the first is to focus on geometry, using the properties of geometric figures combined with algebraic knowledge to investigate. The other focuses on the algebraic aspect, and the geometric properties are only an introduction point, which examines the candidate's computational skills. However, there is no strict distinction between the two types of emphasis, and many of the question types are very similar.

Among them, the construction function of the geometric figure has been given in the figure is the key object of investigation. When doing this kind of question, you must have the main idea of "reducing complexity" and "increasing flexibility".

6. Induction and conjecture of geometric figures.

The high school entrance examination has increased the examination of candidates' ability to induct, summarize and guess, but because the systematic knowledge of the number series will not be formally examined until high school, most of them are placed in the fill-in-the-blank finale questions. For this kind of inductive problem, the method of thinking is the most important.

This problem can be done using the equal area method.

Proof: The perpendicular lines of Ae and DB at point C intersect AE at F and DB at K, respectively, and ace DCB is known by (1).

ae=db, and s ace=s dcb, cf=ck pc bisected apb

In RT PCF and RT PCK, CF=CK, and PC is the common edge.

rt△pcf≌rt△pck

APC = BPC Certification.

f'(x)=3x2-3ax let f'(x)=0 Solution: x=0 and x=a

F(0)=b, f(1)=1-3 2a+b, f(a)=b-1 2a3, f(-1)=-1-3 2a+b

It can be seen that :f(0)>f(a) (because a>1) so :f(0) is the maximum value and f(-1) is the minimum value.

That is, f(0)=b=1, f(-1)=-1-3 2a+b=-2 solution a=4 3

So :f(x)=x3-2x2+1

When m 2 3, the function g(x) has no zero point;

When m=2 3, the function g(x) has and only one zero point;

When 0 m2 3, the function g(x) has two zeros;

When m 0, the function g(x) has and only one zero point;

In summary: when m 2 3, the function g(x) has no zero point;

When m=2 3 or m0, the function g(x) has and only has a zero point;

When 0 m2 3, the function g(x) has two zeros;

3 Analysis: For any b a 0, [f(b)-f(a)] (b-a)<1 is always established, which is equivalent to f(b)-b f(a)-a is always established;

Let h(x)=f(x)-x=lnx+m x-x(x 0), h(x) decrease monotonically on (0,+;

h (x)=1 x-m x 2-1 0 at (0,+ constancy holds, m -x 2+x=-(x-1 2) 2+1 4(x 0), m 1 4;

For m=1 4, h (x)=0 holds only when x=1 2;

The value range of m is [1, 4, +