Urgent! Last National Mathematics Contest Questions!

1. Multiple choice questions (a total of 5 questions, 7 points for each question, full score of 35 points) Each of the following questions is given four options codenamed A, B, C, D, of which there is only one option that is correct, please fill in the code of the correct option in the parentheses after the question, do not fill in, fill in more or wrong 0 points) 1, known non-zero real numbers a, b satisfies |2a-4|+|b+2|+(a-3)b2

4=2a, then a+b is equal to (

a、-1b、0

c, 1d, 2 solutions.

There is a question to know a 3, and the question is set to an equation of |b+2|+(a-3)b20, then a=3, b=-2, so a+b=1, choose c to supplement. 2. As shown in the figure, the side length of the diamond-shaped ABCD is A, the point O is a point on the diagonal AC, and OA=A, ob=oc=OD=1, then A is equal to (

a、5+12

b、5-12

C, 1d, 2 solution: boc abc, b0abbcac is 1a

aa+1a2-a-1=0 Since a 0, the solution is a=5+12 to choose a

I want to ask upstairs, the answer to question 7 is not 60

Aren't 2 7 and 8 10?

The second prize accounts for 2 5+3 4-1=8 20+15 20-1=3 20 first prize 2 5-3 20=8 20-3 20=1 4 third prize 3 4-3 20=15 20-3 20=3 5Please praise Click [Evaluation] in the upper right corner, and then you can select [Satisfied, the problem has been perfectly solved].

Number of second prize winners: (2 5 + 3 4) - 1 = 3 20

1st prize: 2 5-3 20 = 1 5

Third prize: 3 4-3 20 = 3 5

Second Prize: 2 5 + 3 4-1 = 8 20 + 15 20-1 = 3 20

1st prize 2 5-3 20=8 20-3 20=1 4

3rd prize 3 4-3 20=15 20-3 20=3 5

This question is very easy:

I believe that the landlord knows about the similar triangle, assuming that the landlord knows.

In this question:

Assuming that the intersection of DE and AF is H, there are two equations eh: BF=CH:CF=DH:AF

eh:af=ho:of=dh:bf

Simplified to: EH:BF=DH:AF

eh:af=dh:bf

Divide the left and right sides of the two equations at the same time to get it.

bf:af=af:bf

The landlord should have the answer.

Hello, I am a sophomore in the Department of Mathematics at Northeastern University.

Here are two easy ways to give the landlord:

One: From Seva's theorem, (cd da)*(af fb)*(be ec)=1, and cd da=ce eb so af fb=1 so...

Two: The three letters written below together represent the area of the triangle.

af/fb=afo/bfo=afc/bfc=(afc-afo)/(bfc-bfo)=aoc/boc

In the same way, be ec=boa coa, cd da=cob aob, so, (cd da)*(af fb)*(be ec)=1, so...

Certification. In fact, the two methods are the same, and the second method is to prove Seva's theorem.

Seva's theorem is: o is a point in the triangle abc, ao crosses cb, bo crosses ac, co crosses ab in e d f, then (cd da) * (af fb) * (be ec) = 1. Proof refers to solution 2.

2. y x=(y-0) (x-0) represents the slope of the straight line formed by the point on the circle and the origin, so the maximum value of y x is the tangent equation of the circle over the origin, and the tangent equation for the point p(x0,y0) on the circle (x-a) 2+(y-b) 2=r 2 is (x0-a)(x-a)+(yo-b)(y-b)=r 2

1.China is in the 8th East Zone, the Netherlands is in the 1st East Zone, and the time zone difference between China and the Netherlands is 7 hours, and China is 7 hours faster than the Netherlands. Therefore, the first departure was at 4:55 Amsterdam time and landing at 15:15, the whole journey took about 10 hours, and the second take-off was at 21:25 Amsterdam time, and the landing was at 7:10 Amsterdam time the next day, about 10 hours, and there was no contradiction between the two times.

2.In the first quarter, 450kg was normally produced, 90kg was produced overtime, and 20kg was outsourced

In the second quarter, the normal production was 450kg, overtime production was 90kg, and outsourcing production was 200kg

In the third quarter, 750kg was normally produced, 150kg was produced overtime, and 200kg was outsourced

In the fourth quarter, 450kg was normally produced, 90kg was produced overtime, and 110kg was outsourced

Because the cost of overtime production is higher than that of normal production and inventory for a quarter, and the cost of overtime production is lower than that of outsourcing production for a quarter, it is necessary to try to produce normally, work overtime when necessary, and carry out outsourcing production when necessary, and the smaller the inventory period, the better. The first quarter of overtime production and a small amount of outsourcing production, the second quarter and the third quarter to achieve the first grade, can be fully produced, 0 inventory, because the second and third quarters of production has reached the limit, to inventory is actually the outsourcing part of the production will cause cost increases, the fourth quarter of normal production overtime production of a small amount of outsourcing production, to meet the inventory requirements.

3.(1) The capacity of the subway train is a systematic error, which cannot be eliminated and the probability of occurrence is certain, not affected by the peak and non-peak, and the passenger quality is not, because the passengers will automatically look for the team with the least number of people in line, and will spontaneously balance all available channels, only the subway station management work is completely uncontrollable and can be artificially improved.

2) According to the characteristics of different flow directions in the morning and evening peaks, reasonable settings, more inbound gates can be set up in residential areas in the morning, and a large number of outbound gates can be set up in the main commercial areas. Reduce the number of opposing trains, real-time monitoring of the situation, and reasonable planning of different exits.

4.In the picture, the overlord butterfly is horizontally wide, and the singer butterfly is horizontally wide, 5(1) Select a threshold value t=128, the one smaller than the threshold is marked as black, and the one larger than the threshold is extremely white. It is more appropriate to record the number of horizontal rows or vertical rows of black.

2) Take the threshold value t=n, and then take a book k that is larger than the threshold, the smaller than the threshold is recorded as black, and the larger than the threshold is extremely white, and then in these numbers, the larger than k is recorded as true black, and the less than k is recorded as false black, and the ratio r of true and false black is obtained, which can be distinguished.

In addition, the meaning of the question should not allow comparative traits, otherwise it will be obvious at a glance.

The first question: In the bottom picture, the number of numbers in the dark place can be compared to the total number of numbers:

4x5+6x20) (20x20)=7 20 The above solution is wrong, 1) Mathematical language defines mathematical images: such as **, the threshold of the two tables is the same as t=180, then the value greater than or equal to 180 is regarded as white, and the grid chart below 180 is black. The shape of the graph composed of all the black ** can distinguish the corresponding images in Table 1 and Table 2.

Have you done all the five questions? Can you send it... The younger brother is grateful.

1.Solution: 45 (1-15%)=10,000 yuan).

2.Solution: 25000 (1+40%)=35000 (person)3

Solution: The probability of winning the lottery (20 + 30 + 100) 2000 = if all the lottery tickets are issued, at least 200 2000 = 400000 (yuan) The prize amount accounts for at least more sales (20 800 + 30 500 + 100 100) 400000 =

None of your answers are correct.

1) The original surface area of the cube is the sum of the surface areas of 6 squares with sides of 4, i.e. 6x4x4=96

After punching, the surface area of the original large cube will not be reduced, and the small hole does not have an upper one, so there are four more small square surface areas, ie.

4x1x1=4

The surface area of the three-dimensional figure after punching should be 96 + 4 = 100

I don't know where your 110 came from.

2) In the same way, add 4 more small areas on the original basis, that is, 100 + 4 = 104

There's no such thing as 118.

3) No matter how it is expanded, the surface area of the original cube will not be reduced, but the four sides of the small hole will be added.

The area of the four sides is 2(1+x)*1=2+2x

To make the surface area 130, i.e. 100 + 2 + 2x = 130

The solution is x=14

Because it is beyond the edge length of the large cube, it is impossible.

1. The original cube area is 6*4*4=96

After punching, the area is less than 2*1*1=2

At the same time, 4*1*4=16 is added

So in the end it's 96-2+16=110

2. On the basis of the first question, the area is reduced by 4, the area of 1*1 is 4, and 4 of 3*1 area is added, and the area of 3*1 is 12

So the total area is 110-4+12=118

3. The title is 130 square centimeters! The first case is that x is less than 1 then the area cannot reach 130, the second is that x is equal to 1 is not 130, the third is x greater than 1, on the basis of the first problem the area is reduced by 2 x * 1 to 2x, and 4 1 * 1 is reduced to 4, while increasing 2 (x * 4-1) and 2 4 * 1 to 8

So 110-2x-4+2*(x*4-1)+8=1306x=18x=3

So it can be formed.

First, find the surface area of the original cube s1 = 4 * 4 * 6 = 96.

Punch 1 bottom area s2 = 1 * 1 * = 1.

The area of the perforated side is s3 = 1 * 4 * 4 = 16.

1) i.e. s4 = 96 - 2 + 16 = 110

2) The rear punch hole passes through the front hole, so the bottom area of the hole is reduced by 6 (two large positive and negative bodies on the outside, and a small cube with only two sides is formed in the middle of the large cube, and the bottom area of the hole is increased by 4 holes. The two holes coincide). That is, the surface area of an entire hole is increased first, and then the overlapping and decreasing area is subtracted

s5=s1+2*s3-4*s2-6*s2=96+32-10=118

3) The one that passes through is a cuboid. Long xWidth 4, height 1

Based on the result of (1), the area increased is the area of the top and bottom of the cuboid (minus the small holes), and the area of the left and right sides. The area reduced is the side area of the small cube. The area of the front and back sides of the box.

The area of the left and right sides of the cuboid is s6 = 1 * 4 * 2 = 8

The area of the front and back sides of the cuboid is s7 = 1 * x * 2 = 2x

The bottom and top area of the cuboid (excluding small holes) s8=4*x*2-2=8x-2, then the surface area of the mud block s7=s4-4*s2-s7+s6+s8=130x=3