Three practical problems of quadratic equation of one variable in the third grade of junior high school! (30 points can be added) 1. Mr. Li deposits 10000 yuan in the bank. First, he deposits one fixed term for one year. After one year, he automatically transfers the principal and interest to another fixed term for one year. After two years, he gets 10400 yuan of principal and interest. What's the annual interest rate of the deposit? (without considering the interest tax) 2. 200 fruit trees will be planted in an orchard this year. Now it is planned to expand the planting area, so that the planting amount in the next two years will increase by the same percentage as that in the previous year. In this way, the planting amount in three years (including this year) will be 1400, so this percentage can be calculated 3. A container is filled with 20 liters of pure liquid medicine. After several liters are poured out for the first time, it is filled with water. When the same volume of solution is poured out for the second time, only 5 liters of pure liquid medicine are left in the container. How many liters of liquid are poured out each time? Just write the set and the equation, not the solution

Three practical problems of quadratic equation of one variable in the third grade of junior high school! (30 points can be added) 1. Mr. Li deposits 10000 yuan in the bank. First, he deposits one fixed term for one year. After one year, he automatically transfers the principal and interest to another fixed term for one year. After two years, he gets 10400 yuan of principal and interest. What's the annual interest rate of the deposit? (without considering the interest tax) 2. 200 fruit trees will be planted in an orchard this year. Now it is planned to expand the planting area, so that the planting amount in the next two years will increase by the same percentage as that in the previous year. In this way, the planting amount in three years (including this year) will be 1400, so this percentage can be calculated 3. A container is filled with 20 liters of pure liquid medicine. After several liters are poured out for the first time, it is filled with water. When the same volume of solution is poured out for the second time, only 5 liters of pure liquid medicine are left in the container. How many liters of liquid are poured out each time? Just write the set and the equation, not the solution

1. The interest rate of social deposit is X
(1+x)^2=1.04
The solution is x = 0.0198, i.e. 1.98%
2. Let the percentage be X
200*（1+x）^2=1400
The solution is x = 1.6458 = 164.58%
3. Let x liters of liquid be poured out each time
（20-x）*（20-x）/20=5
The solution is x = 10

Another elementary 3 application problem of quadratic equation of one variable (add)
The ratio of length to width of the mirror is 2:1. It is known that the price of the mirror glass is 120 yuan per square meter, and the price of the frame is 30 yuan per meter. In addition, the processing cost of the mirror is 45 yuan. If it costs 195 yuan to make the mirror, how much is the length and width of the mirror?

If the width is x meters, the ratio of length to width is 2:1, so the length is 2x, so the perimeter of the mirror is 2 (2x + x) = 6x,
So the cost of the frame is: 30 × 6x = 180X yuan, the area of the mirror is 2x × x = 2x2, so the cost of the mirror is 120 × 2x2 = 240x2 yuan, and the processing cost is 45 yuan,
So y = 240x2 + 180X + 45,
When y = 195, 195 = 240x ^ 2 + 180X + 45
That is 240x ^ 2 + 180x-150 = 0,
That is, 8x ^ 2 + 6x - 5 = 0,
So X1 = 1 / 2, X2 = - 5 / 4 (rounding off)
Length 2x = 1m
A: the mirror is 1 meter long and 1 / 2 meter wide

A farm planted 10 mu of pumpkin last year, and the yield per mu was 2000 kg. According to the market demand, this year, the farm expanded the planting area, and planted all high-yield new varieties of pumpkin. It is known that the growth rate of pumpkin planting area is twice that of the yield per mu. This year, the total yield of pumpkin is 60000 kg, and the growth rate of pumpkin yield per mu is calculated

Suppose the growth rate of pumpkin yield per mu is x, then the growth rate of planting area is 2x. According to the meaning of the question, we get 10 (1 + 2x) · 2000 (1 + x) = 60000. The solution is: X1 = 0.5, X2 = - 2 (not the meaning, rounding off). Answer: the growth rate of pumpkin yield per mu is 50%

Compare 3333 times 5555 times 6666 times 8888 and 2222 times 4444 times 7777 times 9999

The first formula = the fourth power of 1111 * (3 * 5 * 6 * 8)
The second formula = the fourth power of 1111 * (2 * 4 * 7 * 9)'Is to compare 3 * 5 * 6 * 8 with 2 * 4 * 7 * 9
3*5*6*8＜2*4*7*9
It's very clear······

It is known that the range of F (x) = log2 (x + a) is r, which is the logarithm of 2 (x + a)

The range of F (x) = Log &; (x + a) is r
The value of T = | x | + a can take all positive numbers
Then a ≤ 0

136 × 101-136 simple operation 165 × 77-65 × 77 simple operation

136×101-136=136x(101-1)=136x100=13600
165×77-65×77=(165-65)x77=100x77=7700

For any allowable value of X in a certain range, P = | 1-2x | + | 1-3x | + +|The value of 1-10x | is constant. Find the number P and the range of X
1. For any allowable value of X in a certain range, P = | 1-2x | + | 1-3x | + +|The value of 1-10x | is constant. Find the number P and the range of X
2. It is known that when x takes any value in a certain range, the value of polynomial - 4x + | 4-7x | + | 4-3x | is constant. Try to find out the condition that x should satisfy and find out the constant value

(2)-4x+|4-7x|+|4-3x|=-4x+|7x-4|+|3x-4|
When X4 / 3,
Original formula = - 4x + 7x-4 + 3x-4 = 6x-8

It is proved that when x is very small, 1 / (1 + x ^ 2) is about 1-x ^ 2
How to solve the problem with differential knowledge

(1+x^2)(1-x^2)=1-x^4
When x is very small, x ^ 4 is close to 0
(1+x^2)(1-x^2)=1
So 1 / (1 + x ^ 2) = (1-x ^ 2)

It is known that x = 1 is the solution of the equation 3x with respect to x, and the solution of the equation K with respect to y is - 7K + 5 + Y-1 / 2 = 0

The square of 3x1 - 2x1 + 4x1-7 + k = 0 9-2 + 4-7 + k = 0, so k = - 4
So - 7x (- 4) + 5 + Y-1 / 2 = 0, y = - 65 / 2

What's the result of the quotient of 3.6 divided by 0.24 and the product of 5 times 1.8?