1. Walk into the wonderful world of mathematics

1.9(n- 1)+n = 10n-9 2.630 3。 =36% 4. 133,23 2000=24×53 

5.2520，a=2520n+ 1 6。 A seven. C 8。 B 9。 C 10。 C

1 1.6, two digits 95 must be the divisor of 2003-8= 1995, 1995=3×5×7× 19.

12. 13.

14. Observing the graphic data, we can draw the conclusion that an N-prism has (n+2) faces, 2n vertices and 3n edges.

15.D 16。 A 17。 C S will not decrease with the increase of t, and the distance will not change for a few minutes after repairing the car. After repairing the car, the distance should be increased.

18.C 9+3× 4+2× 4+ 1× 4 = 33。 19.

20.( 1)(80-59)÷59× 100%≈36% (2) 13÷80× 100%≈ 16% 

(3) The growth rate from 1995 to 1996 is (68-59) ÷ 59×100% ≈15%,

The growth rate of other years can be obtained by the same method, and the highest growth rate is 1995 ~ 1996.

2 1.( 1) The list of promotion methods of Mall B is as follows:

Buy more than 24 sets11~ 8 sets of 9 ~ 16 sets 17 ~ 24 sets.

Unit price: 720 yuan, 680 yuan, 640 yuan and 600 yuan.

(2) Comparing the promotion methods of the two shopping malls, we can see that:

Buy 1 ~ 5, 6 ~ 8, 9 ~ 10,1~15.

Select shopping malls b, b, b, b.

Buy 16 sets, 17 ~ 19 sets, 20 ~ 24 sets, more than 24 sets.

Select shopping malls a, b, a and b.

Because it takes 600× 2 1 vcd in mall a and 640× 20 =12800 >12600 to buy 20 vcds in mall b,

So when you buy 20 VCDs, you should go to the mall to buy them.

Therefore, Unit A should go to Shopping Center B, Unit B should go to Shopping Center A, and Unit C should go to Shopping Center A. 。

22.( 1) Arrange divisible rectangles with integer sides from small to large, including

1× 1, 1×2, 1×3, 1×4,2×2, 1×5,2×3,2×4,3×3,2×5,3×4,3×5.

If it can be divided into five eligible papers, because the sum of their areas should be 15, the eligible papers are

1× 1, 1× 2, 1× 3, 1× 4, 1× 5 (as shown in figure ①) or 1× 1.

2. From arithmetic to algebraic answers

1.n2+n = n (n+1) 2.109 3.4.150 minutes 5.c6.d7.b8.b

9.( 1)S = N2(2)① 100② 132-52 = 144(3)n = 15

10.( 1)a & lt； B, (2) If the speed when riding a flat road is "1", then+= (+),

Get =。

1 1.S=4n-4 12。 b2 13.595 14。 ( 1) 18; (2)4n+2

15.a Let the natural number start from a+ 1, and the sum of these 100 consecutive natural numbers is

(a+ 1)+(a+2)+〉…+(a+ 100)= 100 a+5050。

The first column number of 16.c can be expressed as 2m+ 1, and the second column number can be expressed as 5n+ 1.

From 2m+ 1=5n+ 1, n = m, m = 0,5, 10... 1000.

17.A

18.d Hint: Each student moves bricks every hour, and C students move bricks every hour at this speed.

19. Hint: A 1 = 1, A2 =, A3 = ..., An =, original formula =.

20. If each calculator is X yuan, and each book in the math contest handout is Y yuan, then 100(x+3y)=80(x+5y), which means x=5y, then you can buy a calculator = 160 (sets) and a book =800 (volumes).

(2) If it can be divided into six pieces of qualified paper, the sum of its areas should still be 15, but the sum of the areas of the six rectangles above is1×1+1× 2+1× 3+1× 4. 15. Therefore, it is impossible to divide into six pieces of qualified paper.

3. The cornerstone of creation-observation, induction and guessing the answer

1.( 1) 6, (2) 2003.2.A+B = C+D- 14 or a+c=b+d-2 or A+D = B+C3. 13, 3n+.

5.b Tip: The number that appears in these two character strings at the same time is the integer from 1 to 1999 divided by 6, which has 334 * *.

6.C

7. Tip: Observe the written numbers and find that every three consecutive numbers have an even number. Among the first 100 items, 100 is odd, and there are =33 even numbers in the first 99 items.

8. Tip: The arrangement characteristics of this natural number table can be obtained through observation:

(1) Every number in the first column is a complete square number, and it is exactly equal to the square of the number of rows where it is located, that is, the number of the nth row 1 is N2;

② The nth number in the first row is (n-1) 2+1;

③ Line n, decreasing from the first digit to the nth digit1;

④ Increase 1 from the first number in the nth column to the nth number.

It can be found that the number of line 1 0 starting from (1) and column 13 starting from the left should be 10 of column 13, that is, [(13-1]

(2) The number 127 satisfies the relation 127 =112+6 = [(12-1) 2+1]+5, that is.

9.( 1)(2n+ 1)(2n+3)= 4(n+ 1)2- 1；

(2) The number of-rows is 1, 2, 3, ... From 1 to 198, from 1 to 1997 * *, many problems can be easily solved.

10.7N+6，285 1 1。 Forest12.s = 7× 4 (n-1)-5n = 23n-8 (n ≥ 3)13.b65448.

15.( 1) Hint: Yes, the original formula = × 5;

(2) The original formula = there are n- 1 odd numbers in the result.

16.( 1) omitted; (2) number of vertices+number of faces-number of edges = 2; (3) Drawing as required to verify the conclusion of (2).

17.( 1) Generally speaking, we have (a+ 1)+() = = (a+ 1)?

(2) Similar problems such as:

(1) What are two numbers whose difference is equal to their quotient? (2) What kind of three numbers, their sum is equal to their product?

4. Reciprocal and absolute value answers

1.( 1)A； (2)C； (3)D 2。 ( 1)0; (2) 144; (3)3 or -9

3.A = 0，B =。 The original formula =-4.0, 1, 2, …, 1003. Its total is 0.

5.A = 1，B = 2。 Original formula =.

6.a-c 7.m= -x3，n= +x。

∫m =(+x)(+x2- 1)= n[(+x)2-3]= n(N2-3)= n3-3n。

8.P = 3，Q =- 1。 The original formula = 669× 3-(- 1) 2 = 2006.

5. Birds of a feather flock together-the answer to similar items

1. 1 2.( 1)-3, 1 (2)8.3.4000000 4.-4 5.C 6。 C 7。 An eight. A

9.d = > 3 x2-7y+4 y2，F=9x2- 1 1xy+2y2

10. 12 hint: according to the meaning of the question, B = m- 1 = n, C = 2n- 1 = m, 0.625a = 0.25+(-0. 125).

1 1. Right 12. - 13.22

Hint: a>b, original formula = a,

Therefore, it is known that the result of substituting two numbers in each group into algebraic expression is the larger of the two numbers.

Generally speaking, as long as the 50 numbers of 5 1, 52, 53, …, 100 are substituted into each group in turn, the maximum sum of 50 values can be obtained.

15.d16.d17.b18.b Hint: 2+3+…+9+10 = 54,8+9+10.

6. The solution of one-dimensional linear equations

1.- 105.

2. Let the original input number be x, then-1=-0.75, and the solution is x=0.2.

3.- ; 90 4.、- 5.D 6。 A seven. An eight. B

9.( 1) When a≠b, the equation has a unique solution x =；; When a=b, the equation has no solution;

(2) When a≠4, the equation has a unique solution x =；;

When a=4 and b=-8, the equation has countless solutions;

When a=4 and b≦-8, the equation has no solution;

(3) When k≠0 and k≠3, x =；;

When k=0, k≠3, the equation has no solution;

When k=3, the equation has countless solutions.

10. Hint: The original equation is 0x=6a- 12.

(1) When a=2, the equation has countless solutions;

When a≠2, the equation has no solution.

11.10.512.10,26,8, -8 hint: x=, 9-k│ 17, then 9-k =/kloc.

13.2000 prompt: take (+) as a whole. 14. 1.5 15 . a 16 . b 17 . b。

18.d tip: x= integer, and 200 1 = 1× 3× 23× 29, k+ 1.

It can be taken as 1, 3,23,29, (3× 23), (3× 29), (23× 29), 200 1 * * 16, and its corresponding k value is 65438.

19. Children 17, books 150. 20.x=5。

2 1. Hint: Substitute x= 1 into the original equation and get (b+4)k= 13-2a.

This formula applies to any value of k,

That is, there are countless solutions to the equation about k.

Therefore, b+4=0 and 13-2a=0, a= and b =-4 are obtained.

22. Tip: Set the number in the upper left corner of the box to X,

Other numbers in the box can be expressed as:

x+ 1，x+2，x+3，x+7，x+8，x+9，x+ 10，x+ 14，x+ 15，x+ 16，x+ 17，x+2 1，x+22，x+23，x+24，

Judging from the meaning of the question:

X+(x+1)+(x+2)+(x+3)+… x+24 =1998 or 1999 or 2000 or 200 1,

That is, 16x+ 192 = 2000 or 2080.

When the solution is x= 1 13 or 1 18, 16x+ 192=2000 or 2080.

1 13÷7= 16… Yu 1,

That is, 1 13 is the number of 1 in line 17.

The maximum number in this box is113+24 =137; 1 18 ÷ 7 = 16 ... at 6,

That is, 1 18 is the sixth number in line 17.

Therefore, the sum of unbounded numbers in the box is 2080.

7. Solving application problems with equations-interesting answers to travel problems

1. 1 or 3 2.4.8 3.640

4. 16

Tip: suppose that after x minutes, the minute hand and the hour hand overlap for the first time. The minute hand walks 6 degrees per minute and the hour hand walks 0.5 degrees per minute.

Then 6x = 0.5x+90+0.5x5, and the solution is x= 16.

5.C 6。 C prompt: 7. 16

8.( 1) Let the length of CE be x km, then1.6+1+x+1= 2× (3-2× 0.5), and x=0.4 (km).

(2) If the walking route is A→D→C→B→E→A (or A→E→B→C→D→A), the time taken is:

(1.6+1+1.2+0.4+1)+3× 0.5 = 4.1(hour);

If the walking route is A → D → C → E → B → E → A (or A→E→B→E→C→D→A),

Then the time is: (1.6+1+0.4+0.4× 2+1)+3× 0.5 = 3.9 (hours),

Because 4.1>; 4，4 >3.9,

So the walking route should be A→D→C→E→B→E→A (or A→E→B→E→E→C→D→A).

9. Tip: Suppose this person leaves home and the time to start the train is X hours.

From the meaning of the question: 30(x- )= 18(x+), the solution is x= 1,

This person intends to arrive at the train station 10 minutes before the train leaves.

The speed of riding a motorcycle should be =27 (km/h).

10.7.5 prompt: First, find the sum of the speeds of vehicles A and B =20 (m/s).

1 1. 150、200

Tip: Suppose the first car travels (140+x) kilometers,

Then the second car traveled (140+x) × =140+(46+x) km,

X+(46+x)=70。

12.66 13.B

14.d Tip: If the hour hand and the minute hand are at right angles after x minutes, then 6x- x= 180 and x=32.

15. hint: let the train speed be x m/s,

Judging from the meaning of the question: (x- 1)×22=(x-3)×26, x = 14,

Therefore, the train body length is (14- 1)×22=286 (m).

16. If the number of carriage returns is x, the number of departures is (x+6).

When two cars are used at the same time, there is no car in the station.

From the meaning of the question, 4(x+6)=6x+2, and from the solution, x= 1 1.

Therefore, 4(x+6)=68. That is, when the first taxi leaves, the station can't leave on time after at least 68 minutes.

8. Solving application problems with equations-setting the technical answers of elements

1.2857 13

2. Suppose there are X students in this class, and one student is playing football on the playground, 1≤a≤6.

From +a = x, x= a, 3│a,

So, a=3, x=28 (person).

3.24 4.C 5。 B

Tip: Let each cut alloy weigh x grams, and the percentages of copper in 10 kg and 15 kg alloys are respectively

a、b(a≠b)，

Then,

Finishing (b-a)x=6(b-a), so x=6.

6.b prompt: if x cubic meters of gas is used, then 60× 0.8+1.2 (x-60) = 0.88x.

7. Assume that the cost price of each product should be reduced by X yuan.

Then [510× (1-4%)-(400-x) ]× (1+10%) m = (510-400) m gives x =

8. 18、 15、 14、4、8、 10、 1、

9. 1.4 tip: suppose you originally planned to buy X pens and Y ballpoint pens, and the price of ballpoint pens is K yuan.

Then (2kx-ky) × (1+50%) = 2ky+kx, and the solution is y=4x.

10.282.6m tip: set the film width as amm and the length as xmm.

The volume is 0. 15axm3, and the inner and outer diameters of the film wound on the disc are 30mm and 30+015× 600 =120 (mm) respectively. The volume can be expressed as (120-30)? a= 13500a(m3)，

So there is 0. 15ax = 13500a, x = 90000 ≈ 282600, and the film length is about 282600mm, that is, 282.6 mm 。

1 1. 100 Tip: Let the original work efficiency be A and the total workload be B, from -=20, the result is = 100.

12.B 13。 A

14.c prompt: let the purchase price of goods be a yuan and the price be b yuan.

Then 80%b-a=20%a, and the solution is b = a,

Original marked sales profit rate × 100%=50%.

15.( 1)(b-na)x+h

(2) a=2b，h=30b。

If six gates are opened at the same time, the water level relative to the warning line will be (b-na) x+h =-3b after 3 hours.

Therefore, the reservoir can reduce the water level to the warning line within 3 hours.

16.(65438+

Then 2a? T a =a？ T B = T，get T A: T B = 1:2。

(2) From the meaning of the question: =, from (1), T B =2t A,

So, the solution is T=540.

The freight payable to the owner is 540× ×=20=2 160 (yuan).

The car owners of Party B and Party C each get 540×××20 = 4320 (RMB).

9. Line segment answer

1.2a+b 2. 12 3.5a+8 b+9c+8d+5e 4。 D 5。 C

6. A hint: AQ+BC = 2250 & gt；; 1996, so the positions of a, p, q and b are shown as follows:

7.MN & gtAB+NB Tip: Mn = Ma+An = AB, AB+Nb = AB+(CN-BC) = Ab8. Mn = 20 or 40.

9.23 or 1 hint: discuss the dividing point q on the line AP and the dividing point q on the line PB.

10. let AB=x, then the sum of the lengths of the other five sides is 20-x, that is ≤ x < 10.

1 1.3 tip: let AC = X and CB=y, then AD=x+, AB=x+y, CD=, CB = Y, DB=, that is, 3x+ y=23.

12.c Tip: Make five points on the plane, and grasp the hand with connected line segments.

13.d tip: every two n straight lines intersect on the plane, there is at least one intersection point and at most one intersection point.

14. A hint: the maximum information content of each channel is 3+4+6+6= 19.

15.A Tip: The stops are located in Area A, Area B and Area C, and the calculated total distances are 4500m, 5000m and 12000m respectively, so options B and C can be excluded; If the stop is located between area A and area B, and it is x meters away from area A, the total distance is

30x+ 15( 100-x)+ 10(300-x)= 4500+5x & gt； 4500 and exclude option d.

16.( 1) As shown in Figure ①, two straight lines can divide the plane into three or four regions respectively due to their different positions; As shown in Figure ②, three straight lines can divide the plane into four, six and seven regions because of their different positional relationships.

(2) As shown in Figure ③, four straight lines can divide the plane into 1 1 regions at most. At this time, the positional relationship of these four straight lines is intersected by pairwise, and there are no * * * points of three lines.

(3) Divide the plane into areas where n straight lines intersect and no three straight lines intersect at one point. When n= 1, a1+1= 2; When n=2, A2 =1+1+2 = 4; When n=3, A3 =1+1+2+3 = 7; When n=4, A4 =1+1+2+3+4 =11,…

Therefore, the formula an =1+1+2+3+…+n =1+=.

17. Hint: It should be built at the intersection of AC and BC lines.

18. Remember that both banks of the river are L, L' (pictured). If the straight line L is translated to the position of L', the point A will be translated to A', the connecting rod A'B will intersect L' at D, and if the bridge intersects D, it will be at CD.

10. corner answer

1.45 2.22.5 hint: 15× 6- 135× 0.5.

3. 15 4.6 5.B 6。 A seven. C 8。 B

9.∠COD=∠DOE Tip: ∠ AOB+∠ DOE = ∠ BOC+∠ COD = 90.

10.( 1) The following schematic diagram is for reference only.

(2) Omission

1 1.345 hint: because 90

So 6

So α+β+γ = 23x15 = 345.

12.∠EOF、∠BOD、BOC∠BOF,∠EOC

13. If the light is within ∠AOB, ∠ AOC = 8 20'; If the light OC is outside ∠ AOB, ∠ AOC =1514.4015.c16.d.

17.20 hint: it is very simple to solve this problem with equations.

Let ∠ COD = X, ∠ BOC+∠ AOD = Y, from the meaning of the question:

18. Tip: * * * There are four times when the angle between the hour hand and the minute hand is 60.

(1) The first time was exactly two o'clock.

(2) When the second time is set to 2 o'clock X, and the angle between the hour hand and the minute hand is 60, then X = x= 10++10+00, and the result is x=2 1.

(3) If the angle between the hour hand and the minute hand is 60, then y+ 10=+15, and the result is y=5.

(4) For the fourth time, if the angle between the hour hand and the minute hand is 60, then Z = z= 15++10+00, and the solution is z=27.

19. Tip: If you only use the template continuously, you will get an integer multiple of 19, that is, you can draw 19 corners continuously with the template to get 36 1 corners, and remove the rounded corners of 360 to get 65438+.