The master of junior high school math questions is coming, a junior high school math problem, seekin

Hehe, that's how it works, remember to give me extra points after reading it. If you don't understand, send me a message.

a^5+4a

a^5-5a^3+4a+5a^3

a(a^4-5a^2+4)+5a^3

a(a^2-1)(a^2-4)+5a^3

a(a+1)(a-1)(a+2)(a-2)+5a^3

So a 5 + b 5 + c 5 + 4 (a + b + c).

a^5+4a+b^5+4b+c^5+4c

a(a+1)(a-1)(a+2)(a-2)+5a^3+b(b+1)(b-1)(b+2)(b-2)+5b^3+c(c+1)(c-1)(c+2)(c-2)+5c^3

a(a+1)(a-1)(a+2)(a-2)+b(b+1)(b-1)(b+2)(b-2)+c(c+1)(c-1)(c+2)(c-2)+5(a^3+b^3+c^3)

From the nature of the number, if a is any integer, then a(a+1)(a-1)(a+2)(a-2) represents five consecutive integers.

Of the five consecutive integers, one must be a multiple of 5, then a(a+1)(a-1)(a+2)(a-2) is a multiple of 5.

If at least 1 of the 5 consecutive integers is a multiple of 4 and at least 1 is a multiple of 2, then a(a+1)(a-1)(a+2)(a-2) is a multiple of 8.

If at least 1 of the 5 consecutive integers is a multiple of 3, then a(a+1)(a-1)(a+2)(a-2) is a multiple of 3.

So a(a+1)(a-1)(a+2)(a-2) must be a multiple of 5 8 3 120.

In the same way, b(b+1)(b-1)(b+2)(b-2) and c(c+1)(c-1)(c+2)(c-2) must also be multiples of 120.

So, if a 5 + b 5 + c 5 + 4 (a + b + c).

a(a+1)(a-1)(a+2)(a-2)+b(b+1)(b-1)(b+2)(b-2)+c(c+1)(c-1)(c+2)(c-2)+5(a^3+b^3+c^3)

is a multiple of 120, then 5 (a 3 + b 3 + c 3) must be a multiple of 120.

So a 3 + b 3 + c 3 must be a multiple of 24.

1) A 5+b 5+c 5+4 a+b+c is a multiple of 120?

Look at a 5+4a to prove that it is divisible by 3,5,8 (because 120=3*5*8).

Divisible by 3: a = 3k (k is an integer), a 5 divided by 3 and 1 when a = 3k + 1, 4a divided by 3 and 1

I can't do it anymore.

The title is wrong! A = 4, b = 3, c = 3 can be substituted to get a 5 + b 5 + c 5 + 4 a + b + c = 1550, not divisible by 5.

2) Verify that a3+b 3+c 3 is a multiple of 24?

The title is wrong! It can be substituted for a = 4, b = 3, c = 3 to get a 3 + b 3 + c 3 = 118, which is not divisible by 24.

It can be proved that a 5 + b 5 + c 5 + 4 a + b + c minus a 3 + b 3 + c 3 by a factor of 5

Is it 0? After the equation, the common factor is extracted, and three formulas are divided into a, b, and c, and the factor is decomposed. a(a 2-1)(a 2-4) +b(b 2-1)(b 2-4) + c(c 2-1)(c 2-4) = 0 abc can be taken as an integer, how to prove it next......Let's make up the numbers for yourself.

It's time to add points upstairs, the method has already been told to you. Club!!!

Consumption amount x range (RMB) 200 a<400 400 a<500 500 a<700 700 a 900 The amount of lottery tickets (RMB) 30 60 100 130 (1) For goods with a price between 500 yuan and 800 yuan (including 500 yuan and 800 yuan), how much can customers get a one-third discount rate for the goods with a price tag? Solution: (1) Preferential amount:

1000 (1-80%)+130=330 (yuan) preferential rate: 3301000 100%=33%; 2) Set the purchase price of a product with a price of x yuan to get a preferential rate of 13 When buying a product with a price tag between 500 yuan and 800 yuan, the consumption amount a is between 400 yuan and 640 yuan When 400 a 500, 500 x 625 From the title, get: 13x solution:

x=450 but 450 500 is not in line with the topic, so it is discarded; When 500 A 640, 625 x 800 is derived from the question: 13x Solution: x=750 and 625 750 800, which is in line with the question Answer:

Buy an item with a price tag of 750 yuan and get a discount rate of 13.

Since the consumption amount of the goods listed within [500,800] (yuan) satisfies the consumption amount of 400, it is necessary to discuss the range of consumption amount (yuan), and then solve the inequality group to obtain the answer to the problem

Solution: Set the price of the product to be X yuan

then 500 x 800, consumption: 400

From known (1).

13500 x 800, or (2).

13500 x 800, the inequality group (1) has no solution, the solution of the inequality group (2) is: x=750 Therefore, when the customer buys the list price of 750 yuan, he can get a discount rate of 13 Gone.

1. The sum of the above two angles and the angle BAC is 90 degrees, so the two angles are equal.

2. BF=AC, CG=AB and their angles are equal, so the triangle ACG is equal to ABF, so AF=AG

3. Use the corner edges to find the congruence of bef and cef, so bf=ce

The first problem uses a right triangle to prove that one acute angle is equal to the right angle minus another acute angle.

The second question makes use of corner edge congruence. The third edge is equal.

The third problem uses the congruence of corners and edges to obtain bf=ce

Solution: according to the preferential plan (1) purchase, payable: 200 20 + (x-20) 40 = 40x + 3200 (yuan), according to the preferential plan (2) purchase, payable: (200 20 + 40x) 90% = 36x + 3600 (yuan), let y = (40x + 3200) - (36x + 3600) = 4x-400 (yuan), when y 0, that is, (20 x 100 and an integer) when option (1) is more cost-effective than plan (2), when y = 0, That is, when x = 100, choose two schemes to save money, when y 0, that is, (x 100 and an integer) when option (2) is more cost-effective than scheme (1), if you choose scheme (1) and scheme (2) at the same time, then in order to get the number of the factory gift tie at the same time enjoy a 9% discount, you can consider designing another scheme (3), that is:

First buy 20 suits according to (1) plan and get 20 ties, and then the remaining (x-20) ties are purchased according to the preferential plan (2), payable: 200 20 + (x-20) 40 90% = 36x + 3280 (yuan) plan (3) compared with plan (2), obviously plan (3) is more cost-effective, plan (3) is compared with plan (1), when 36x+3280 40x+3200 solution x 20, that is, when x 20 Option (3) is more cost-effective than option (1).

To sum up, when x 20, plan (3) is the most cost-effective

Hello! Now let's discuss the budgets for the two purchase options under the headings.

1. Total amount = 200 * 20 + 40 * (x-20) = 40x + 3200;

2. Total amount = (200 * 20 + 40x) * 90 = 36x + 3600;

Step 3: Let 40x + 3200=36x +3600 to get the value of x, x =100

Conclusion: The premise of the question is x>20, when x<100, choose option 1; When x>100, select option 2.

One plan: 20x200 + (30-20) x 40 = 4400 yuan, two plans: yuan.

So the first option is more cost-effective than the second option.

Answer: (1) The length of CQ, this understanding is different, I understand that it is 0 after exercise, that is, 0T, if it is CQ length before movement is 2T, T=3;

2) Question 1: According to the requirements of the two areas to be equal, the area formula is the base multiplied by the height divided by 2, that is, the height is the same as 6, the requirement is ap=qf, that is, ap 2=qc, the speed of the two lines is different, the equation can be listed: 1t 2=6-2t, get t = seconds;

1) Solution: Set up X A-type cars, which can be installed with Y units.

According to the title: xy=270

y+15)•(x-2)+30=270

Solution: y=15x divided by 2

Handful. Bring in 15x square = 540

The solution is x=6

Bring x=6 to get y=45

45 + 15 = 60 (sets).

A: 45 units for type A and 60 units for type B.

2) Solution: Set up X cars of type A and X + 1 cars of type B, according to the question: 45X+60 (X+1)=270

The solution is x=2

2+1=3 (vehicles).

2x350 + 3x400 = 1900 (yuan).

Answer: 2 A-type cars and 3 B-type cars are required, a total of 1900 yuan.