Solutions to practical problems of quadratic equation of two variables In the car with constant speed, Xiao Liang noticed that the number on the road milestone was two digits. One hour later, he saw that the number on the milestone and the first two digits were in reverse order. Another hour later, he saw that the number on the milestone for the third time was just the first two digits. A three digit zero was added between the two digits. What were the numbers on the three milestones?

Suppose the first milestone has a single digit number of B and a ten digit number of a, then the first milestone is 10A + B
The second milestone is 10B + a
The third milestone is 100A + B
Because the speed is equal and the time interval is equal, so
The distance between the first and second milestones = the distance between the second and third milestones
10B + A - (10A+B) = 100A+B - (10B+A)
9B-9A = 99A-9B
108A = 18B
6A = B
Because a and B are all numbers, and a is the ten digit number of the first milestone, and B is the ten digit number of the second milestone
So the values of a and B can only be 1,2,3,4,5,6,7,8,9
So obviously a = 1, B = 6
So first milestone 16, second milestone 61, third milestone 106
They are all 45 meters apart

Ask some questions about the application of quadratic equation of two variables
One hundred and fifty students of a certain school take part in the mathematics examination, with an average of 55 points per person. Among them, 77 points are passed and 47 points are failed. How many students are passed or failed?
A cage contains chickens and rabbits, which have 8 heads and 22 feet. How many chickens and rabbits are there in each cage?
A school bought seven color TV sets of two types, which cost 15900 yuan. It is known that the two types of color TV sets are 3000 yuan and 1300 yuan respectively. How many color TV sets do you buy?
If you plant trees along a road every 3 meters, there are still three at the end. If you plant trees every 2.5 meters, there are still 77 at the end. How long is the road? How many trees are there?
There is a large flock of sheep, some of which have already gone up the mountain, and the other part is still at the foot of the mountain. If three of the sheep at the foot of the mountain go up the mountain, the whole sheep at the foot of the mountain are one third of the sheep. If three sheep come down from the mountain, there will be as many sheep at the foot of the mountain. Do you know how many sheep there are at the foot of the mountain?

(1) Let x pass and Y fail
x+y=150,
77x+47y=55×150
x=40,y=110.
(2) Let chicken and rabbit be x, y respectively,
x+y=8,
2x+4y=22
x=5,y=3.
(3) Let a and B buy X and Y sets respectively,
x+y=7,
3000x+1300y=15900
x=4,y=3.
(4) There are x meters on the road and Y trees,
x÷3+3=y,
x÷2.5-77=y,
x=1200,y=403.
(5) Suppose there are x heads on the mountain and Y heads at the foot of the mountain,
x+3=2（y-3）
x-3=y+3,
x=21,y=15.

Using the square of factorization (2x + 1) - 16 = 0, one of them is 2x + 5 equals 0. What's the other equation of first degree? (I've been thinking about it for a long time,

It is known that the eccentricity of the ellipse C: x2 / A2 + Y2 / B2 = 1 (a > b > 0) is √ 6 / 3, and the distance from one end of the minor axis to the right focus is √ 3
2）
Let the line L and the ellipse C intersect at two points ab. the distance from the coordinate origin o to the line √ 3 / 2 is the maximum area of the triangle AOB

According to the meaning of the question, e = C / a = √ 6 / 3 and B ^ 2 + C ^ 2 = (√ 3) ^ 2 = a ^ 2
The simultaneous solution is a ^ 2 = 3, C ^ 2 = 2, B ^ 2 = 1
That is, the elliptic equation is x ^ 2 / 3 + y ^ 2 = 1
Let L: y = KX + B be a straight line
Because: the distance d from the coordinate origin o to the line L is √ 3 / 2
Then the formula of distance from point to line is obtained
d=√3/2=|b|/√[k^2+1]
Then: B ^ 2 = (3 / 4) (k ^ 2 + 1)
Because: the intersection of line L and ellipse C and two points a and B
Let a (x1, Y1) B (X2, Y2)
According to the chord length formula of the intersection of straight line and ellipse, the following formula is obtained:
|AB|
=√[k^2+1]*|x1-x2|
=√[k^2+1]*√[(x1+x2)^2-4x1x2]
because:
Ellipse C: x ^ 2 / 3 + y ^ 2 = 1
Straight line L: y = KX + B
Then, LIANLI can obtain:
x^2/3+(kx+b)^2=1
[(1+3k^2)/3]x^2+2kbx+b^2-1=0
Because a and B are the intersection points,
Then X1 and X2 are two parts of the equation
Then from the Weida theorem, we get that:
x1+x2=-6kb/(1+3k^2)
x1x2=(9k^2-3)/(12k^2+4)
Then:
|Ab | = √ [K ^ 2 + 1] * √ [(x1 + x2) ^ 2-4x1x2]
=√{3+4/(3k^2+1)-4/[(3k^2+1)^2]}
Let t = 1 / (3K ^ 2 + 1) (t belongs to (0,1))
|AB|=√[3+4t-4t^2]
=√[-4(t-1/2)^2+4]
Then when t = 1 / 2, | ab | takes the maximum value = 2
In this case, k = ± √ 3 / 3
Maximum area of △ AOB
=(1 / 2) | ab | Max * D
=(1/2)*2*(√3/2)
=√3/2

1-1 / 2,1 / 2-1 / 3,1 / 3-1 / 4,1 / 4-1 / 5. Write its general formula

1/n-1/（n+1）

Three x minus two brackets, x plus two x minus one bracket, where x equals minus two

3x²-2(x²+2x-1)
=3x²-2x²-4x+2
=x²-4x+2
=4+8+2
=14

A fraction by three Sevens by fifteen is one

We know that x times the N-1 power of x times the 2n power of x = the 15th power of X
The (1 + n-1 + 2n) power of x = the 15th power of X
1+n-1+2n=15
3n=15
N = 52 times n to the second power - n
=The second power of 2x5-5
=2X25-5
=50-5
=45

IF(A3="","",A3),

If there is no data in A3, there is no data in this cell, otherwise it is equal to A3
It's actually copied here

The solution of quadratic inequality with one variable and the equation with absolute value sign and the solution of linear inequality with one variable
I don't know how to solve these problems. I hope you can teach me one by one. Thank you

Main knowledge: 1. Geometric meaning of absolute value: it is the distance from the point on the exponential axis to the origin; it is the distance between two points on the exponential axis. 2. The solution of inequality of type and. At that time, the solution set of inequality is inequality

80 hectares is equal to several parts of a square kilometer
The simplest

1 hectare = 10000 square meters = 0.01 square kilometers
80 ha = 0.8 km2 = 4 / 5 km2