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Looking for 100 math problems in the first semester of junior high school (with answers)
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Junior one olympiad exercises 1

A consumption 100 yuan, debt 600 yuan after three years. What is everyone's annual income?

What is the sum of the last four digits of s?

4. A person goes uphill at a speed of 3 km/h and downhill at a speed of 6 km/h, and the journey 12 km * * * It takes 3 hours and 20 minutes to try to find the distance between uphill and downhill.

5. Total:

6. Prove that the remainder obtained by dividing the prime number p by 30 must not be a composite number.

8. If two integers X and Y make x2+xy+y2 divisible by 9, it is proved that X and Y are divisible by 3.

9. As shown in figure 1-95. In quadrilateral ABCD, the midpoint of diagonal AC and BD is m, n, and the extension line of MN intersects with AB side at point P, which proves that the area of △PCD is equal to half the area of quadrilateral ABCD.

Answer:

So x=5000 yuan.

So the sum of the last four digits of S is 1+9+9+5 = 24.

3. Because

A-b≥0, that is, a ≥ B. that is, when B.

When ≥ a > 0 or b ≤ a < 0, the equation holds.

4. Set an uphill journey of X kilometers and a downhill journey of Y kilometers.

have

From ② 2x+y=20, ③

If ① has y = 12-x, substitute ③ to get 2x+ 12-x = 20.

So x=8 (km), so y=4 (km).

5. item n is

therefore

6. Let p = 30Q+R, 0 ≤ R < 30. Because p is a prime number, r≠0, that is, 0 < R < 30. Suppose that R is a composite number, because R < 30, the minimum prime number of R can only be 2, 3, 5. Then, from P = 30Q+R, when

7. Establish

(2p- 1)(2q- 1)=mpq of formula ①, i.e.

(4-m)pq+ 1=2(p+q)。

It is known that m < 4. From ①, m > 0 and is an integer, so m= 1, 2,3. Let's study p and q separately.

(1) If m= 1, there is

The solutions are p= 1 and q= 1, which are inconsistent with the known ones and are discarded.

(2) If m=2, there is

Because 2p- 1=2q or 2q- 1=2p is impossible, there is no solution when m=2.

(3) If m=3, there is

Get a solution

So p+q = 8.

8. Because x2+xy+y2 = (x-y) 2+3|xy .. Topic setting, 9 | (x2+xy+y2), so 3 | (x-y) 2. Because 3 is a prime number, 3 | (x-y). If 3 | y, you can also get 3 | x.

9. Connect AN and CN, as shown in figure 1- 103. Because n is the midpoint of BD, so

The above two formulas add up.

On the other hand,

S△PCD=S△CND+S△CNP+S△DNP。

So prove it to me.

S△AND=S△CNP+S△DNP。

Because m and n are the midpoint of AC and BD, respectively,

S△CNP=S△CPM-S△CMN

=S△APM-S△AMN

=S△ANP。

And S△DNP=S△BNP, so

S△CNP+S△DNP = S△ANP+S△BNP = S△ANB = S△AND。

Exercise 2 of Olympiad Mathematics in the first day of junior high school

1. Given 3x2-x= 1, find the value of 6x3+7x2-5x+2000.

2. A store sells 100 items a day, and each item can make a profit in 4 yuan. Now they increase profits by raising prices and reducing purchases. According to experience, every time the price of this commodity increases by 1 yuan, it will sell less 10 pieces every day. How much can I raise per piece to get the maximum profit? What is the maximum profit?

3. As shown in figure 1-96, it is known that CB⊥AB, CE ∠BCD, DE ∠CDA, ∠ 1+∠ 2 = 90. Verification: Da ⊥ Company.

4. Known equation

The solution should be

A student copied C wrong when solving a problem, so the solution is

Find the value of A2+B2+C2.

5. Find the integer solution of the equation | xy |-| 2x |+y | = 4.

6. Wang Ping bought a three-year treasury bill with an annual interest rate of 7. 1 1% and a five-year treasury bill with an annual interest rate of 7.86% * * 35,000 yuan. If the three-year Treasury bill matures, he will deposit the principal and interest into two consecutive one-year time deposits. Five years later, the total principal and interest with the five-year treasury bonds is 477,665,438+0 yuan. He asked Wang Ping to buy three-year treasury bonds. (The annual interest rate of one-year time deposit is 5.22%)

7. For what values of k and m, the equations have at least one set of solutions?

8. Find the integer solution of the indefinite equation 3x+4y+ 13z = 57.

Xiao Wang bought 40 fruits for five friends with 5 yuan money. There are three kinds of fruits: apples, pears and apricots, each with 20 points, 8 points and 3 points respectively. Xiao Wang hopes that he and his five friends can get apples. Everyone gets a different number of apples. Can he achieve his wish?

Answer:

1. The original formula = 2x (3x2-x)+3 (3x2-x)-2x+2000 = 2x ×1+3x1-2x+2000 = 2003.

2. It turns out that you can make a profit of 4× 100 yuan every day. If the price of each commodity is increased by X yuan, the profit of each commodity is (4+X) yuan, but (100- 10x) pieces are sold every day. If the daily profit is set to y yuan, then

y =(4+x)( 100- 10x)= 400+ 100 x-40x- 10 x2 =- 10(x2-6x+9)+90+400 =- 10(x-3)2+490。

So when x=3, y = the maximum value of 490 yuan, that is, 3 yuan, increases the price of each item, and gains the most every day, and 490 yuan.

3. Because CE bisects ∠BCD, DE bisects ∠ADC, ∠ 1+∠ 2 = 90 (Figure 1- 104), so

∠ADC+∠BCD= 180,

So ad ∨ BC. ① Because of ab⊥bc②.

Advertised by ①, ② ab ⊥.

According to the meaning of the question, there are

So A2+B2+C2 = 34.

5. | x |||| y |-2 | x| +| y | = 4, that is | x | (| y |-2)+(| y |-2) = 2,

So (| x |+ 1) (| y |-2) = 2.

Because | x |+ 1 > 0, and both x and y are integers, so

So there is

6. Suppose Wang Ping buys three-year and five-year government bonds with X yuan and Y yuan respectively, then

Because y=35000-x,

So x (1+0.0711× 3) (1+0.0522) 2+(35000-x) (1+0.0786× 5) = 4777.

So1.3433x+48755-1.393x = 47761,

So 0.0497x=994,

So x=20000 yuan, y=35000-20000= 15000 yuan.

7. Because (k- 1) x = m-4, ①

When m is a real number, the equations have a unique solution. When k= 1 and m=4, the solutions of ① are all real numbers, so the equations have infinite solutions.

When k= 1, m≠4, ① there is no solution.

Therefore, when k≠ 1, m is any real number, or k= 1, m=4, the equations have at least one set of solutions.

8. From the problem set equation

z=3m-y。

x = 19-y-4(3m-y)-m = 19+3y- 13m。

The general solution of the original equation is that n and m take arbitrary integer values.

9. Suppose that apples, pears and apricots are bought by X, Y and Z respectively, then

If y is removed, we get 12x-5z = 180. Its solution is x=90-5t and z = 180- 12t.

Substituting into the original equation, we get y=-230+ 17t, so x=90-5t, y =-230+ 17t, z = 180- 12t.

x=20,y=8,z= 12。

So Xiao Wang's wish can't come true, because according to his request, there must be at least1+2+3+4+5+6 = 21> 20 apples.

Exercise 3 of Olympiad Mathematics in Grade One of Junior High School

1. Solve the equation about x.

2. Solve the equation

Where a+b+c ≠ 0.

3. Find the sum of the coefficients in the expansion of (8x3-6x2+4x-7)3(2x5-3)2.

4. Pour out 8 liters of liquid pesticide and fill it with water, then pour out 4 liters of mixed solution and fill it with water. At this time, the pesticide concentration was 72%. Find the capacity of the barrel.

5. How many natural numbers x*** satisfy [- 1.77x]=-2x? Here [x] represents the largest integer not exceeding x, for example, [-5.6]=-6, [3] = 3.

6. Let p be a point in △ABC. Find the distance range from p to △ABC and the ratio to the perimeter of triangle.

7. Party A and Party B walk in opposite directions from the East and West Stations at the same time. When they meet, Party A travels 24 kilometers more than Party B, and it takes 9 hours for Party A to arrive at the East Station and 16 hours for Party B to arrive at the West Station. Find the distance between the two stops.

8. There are three numbers written on the blackboard. Erase one of them at will and rewrite it as the sum of the other two numbers minus 1. Continue like this, and finally get 19, 1997, 1999. Can the original three numbers be 2, 2 and 2?

9. There are n real numbers x 1, x2, …, xn, and each real number is either+1 or-1, and

Prove that n is a multiple of 4.

Answer:

1. Simplified to 6(a- 1)x=3-6b+4ab, when a≠ 1

2. Transform the original equation into

X = a+b+c can be obtained from this.

3. When x= 1, (8-6+4-7) 3 (2-1) 2 =1. That is to say, the sum of the coefficients in the expansion is 1.

According to the meaning of the question

Denominator removal and simplification, the result is 7x2-300x+800=0, that is, 7x-20)(x-40)=0.

5. If n is an integer, then [n+x] = n+[x], so [-1.77x] = [-2x+0.23x] =-2x+[0.23x].

From the known [- 1.77x]=-2x, so -2x=-2x+[0.23x], so [0.23x] = 0.

And because x is a natural number, 0 ≤ 0.23x < 1. Experiments show that x can be 1, 2, 3, 4, ***4.

6. As shown in figure 1- 105, there is BC < Pb+PC in △PBC, ①.

Extend BP to AC to D. It is easy to prove that Pb+PC < AB+AC. ②

From ①, ② BC < Pb+PC < AB+AC, ③

Similarly AC < PA+PC < AC+BC, ④

AB