A slightly difficult math problem 1st grade of junior high school .

Set up in A did the X quarter.

Total salary of A: set up in B for y years.

Total salary of B: 6+8+...6+2(y-1)]

6y+y(y-1)/2*2

y^2+5y

Because there are 4 quarters in 1 year, x = 4y

Total salary of A: =>y 2+5y

A has a higher salary than B.

Your answer is wrong, if you do it for 1 year, A has 10,000 yuan) > B has 60,000 yuan.

It can be seen that Company B was high in the first 6 years, and Company A was high after 6 years.

There is a problem with this question, the salary of Company A in the first year is 4*00,000 yuan, and it can only increase by 10,000 yuan every year in the future, while Company B increases by 20,000 yuan per year, which is always more than that of Company A.

If you do it for 1 year, A has 10,000 yuan) > B has 60,000 yuan.

It can be seen that Company B was high in the first 6 years, and Company A was high after 6 years.

I don't think I have learned such a complex topic as X in the first year of junior high school.

Your answer is wrong! Sometimes the answer isn't necessarily right!

Quiz everyone: This is a math problem that can measure whether a person has business acumen. Master Wang is selling fish, a pound of fish at a price of 45 yuan, now a big sale at a loss, the customer bought a kilogram for 35 yuan, gave Master Wang 100 yuan fake money, Master Wang has no change, so he asked a neighbor for 100 yuan.

Afterwards, the neighbor found out that the money was fake in the process of depositing money, and it was confiscated by the bank, and Master Wang lost 100 yuan to the neighbor, how much did Master Wang lose in total?

Note: The difference between catties and kilograms.

A total loss of 100 + (45 2-35) = 100 + 55 = 155 yuan.

The fish pays 55, the neighbor pays 100, and the customer pays 65, for a total of 220

The circle is 2, the box is 1, and the triangle is 5

The big ones are added on the inside and the big ones are subtracted on the outside.

The last one is: 5-2=3

You should fill in: 3 is really difficult for the first grade.

I don't know what to do. Write 2 well, reason: two shapes, one circle, one triangle.

Countercurrent velocity = hydrostatic velocity – water-air velocity.

Downstream velocity = hydrostatic velocity + water velocity.

It's OK to figure this out.

From the moment the dinghy falls overboard, the speed at which the dinghy drifts is equal to the speed of the water.

The sum of the speed of a rubber boat and a steamer is equal to the speed of the steamer in the early book of the steady.

After the ship turns around, the difference in speed between the ship and the rubber boat is also equal to the hydrostatic speed of the ship.

Therefore, the time it takes for the rubber boat to drift before the ship turns around is equal to the time it takes for the ship to catch up with the rubber boat after the ship turns around.

The time when the rubber boat falls into the water is 5:00-1:00=4:00 in the afternoon

If the water is used as a reference, this problem is easier to solve, that is, the water is always stationary, then the speed of the boat back and forth is unchanged, at 5 o'clock in the afternoon, the rubber boat was found to fall into the water, and after an hour of reading, then the rubber boat was dropped 1 hour ago, so the rubber boat was lost at 4 o'clock in the afternoon. Aberdeen next to the head.

1. Because a 2 + a + 1 = 0, a 2 + a = -1.

a^4+a^3-3a^2-4a+3

a^2*(a^2+a)-3a^2-4a+3=a^2*(-1)-3a^2-4a+3

a^2-3a^2-4a+3

4a^2-4a+3

4(a^2+a)+3

63. (1) This question is very simple.

Both sides of the time-sharing equation are multiplied by x 2-1 at the same time

Get: x 2-1-7(x-1) = 14

x^2-7x-8=0

x-8)(x+1)=0

So: x=8

x=-1 (tested to be rooted, rounded).

So the solution of the original equation is x=8

2) Multiply both sides of the equation by 2x (x-7) at the same time

Get: (2x) 2+(x-7)2=4x(x-7) Simplify: x 2+14x+49=0

x+7)^2=0

x=-7 is the solution of the original equation after testing.

There are problems with the next few questions, and I can't understand which ones are exponents.

4、b^(2m+2)÷(b^2÷b^2m)÷(b^2m)^2=b^(2m+2)÷[b^(2-2m)]÷b^4m)=b^[(2m+2)-(2-2m)-4m]=b^0

1 (when b is not equal to 0, it does not make sense when it is equal to 0) 5(2xy)^3÷(6x^2y^3*1/3xy^(-2))=8x^3y^3÷(2x^3y)

4y^26..6x^(n+1)+5x^n*y-4x^(n-1)*y^2=x^(n-1)(6x^2+5xy-4y^)=x^(n-1)(3x+4y)(2x-y)

This involves the idea of "overall substitution" in mathematics.

Because a 2+a+1=0, a 2+a=-1.

a^4+a^3-3a^2-4a+3=a^2*(a^2+a)-3a^2-4a+3

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

a^2-3a^2-4a+3

4a^2-4a+3

4(a^2+a)+3

The box is the sum of the numbers in the two boxes connected to it on the right;

Answers: 6, 4, 1; 4,2,1