Notes on Sun Tai in classical Chinese Not too long

Notes on Sun Tai in classical Chinese Not too long

Sun Tai is a native of Shanyang. He learned from Huang Fuying when he was young. His ambition and moral character are very ancient. Sun Tai's wife is his aunt's daughter. At first, his aunt was old and entrusted his two daughters to sun Tai, saying, "if one eye of the eldest daughter is wrong, you can marry her sister." his aunt died, and sun Tai married his aunt's eldest daughter as

Two mathematical problems about the equation of higher degree (including parameter) in one variable,
1. When m is an integer, the equation (m-1) x ^ 2 - (2m + 1) x + 1 = 0 has an integer root
2. Solve the equation: 5x ^ 2 + x-x * √ (5x ^ 2-1) - 2 = 0

1. If M is not equal to 1, then B ^ 2-4ac = 4m ^ 2 + 5 > 0. So the equation has two real roots X1 and x2. If at least one root of X1 and X2 is an integer root, then there is an integer k such that 4m ^ 2 + 5 = k ^ 2, that is, the discriminant can be square root.. that is, (2m) ^ 2 + 5 = k ^ 2. Because 2m must be even, then K is odd. Only m = ± 1, k = 3 is suitable
Because m ≠ 1, M = - 1
2. Let t = √ (5x ^ 2 - 1) ≥ 0
Then 5x ^ 2 = T ^ 2 + 1
The original equation is T ^ 2 + 1 + X - XT - 2 = 0
t^2 - 1 - xt +x =0
(t+1)(t-1) - x(t-1) =0
(t-1)(t+1-x) =0
T = 1 or T = X-1
T = 1, that is √ (5x ^ 2 - 1) = 1
The solution is x = ± √ 10 / 5
T = X-1, that is √ (5x ^ 2 - 1) = X-1,
We get 2x ^ 2 + X-1 = 0
The solution is x = 0.5, x = - 1
Because both 0.5 and - 1 are less than 1
Here t = X-1

One variable quadratic equation!
100+100（1+x）+100(1+x)²=331
This is an application problem... Suddenly short circuit, do not know how to solve this equation~

100+100（1+x）+100(1+x)²=331
100+100（1+x)+100(1+2x+x²)=331
100+100+100x+100+200x+100x²=331
100+100+100x+100+200x+100x²-331=0
100x²+300x-31=0
∵a=100,b=300,c=-31
b²-4ac=300²-4×100×(-31)
=90000+12400
=102400
∴x=(-300±√102400)/(2×100)
=(-300±320)/200
x1=0.1,x2=-3.1

Help solve the first three of the two equations of one variable! Online wait for urgent!
3（2-y)^2=y(y-2)
2(x+1)^2=x^2-1
x^2+4x-1=0
x^2+3x+2=0
x^2+4x+4=0

3(2-y)^2=y(y-2)
(y-2)(3y-6-y)=0
Y = 2 or y = 3
2(x+1)^2=x^2-1
(x+1)(2x+2-x+1)=0
X = - 1 or x = - 3
x^2+4x-1=0
(x+2)^2=5
Then x + 2 = radical 5 or - radical 5
X = radical 5-2 or - radical 5-2
x^2+3x+2=0
(x+1)(x+2)=0
X = - 1 or x = - 2
x^2+4x+4=0
(x+2)^2=0
Then x = - 2

In the third year of junior high school, the quadratic equation of one variable is solved,
(20-2x)*(12-2x)=203.04

The simplification is 240 -- 64x + 4x square = 203.4, and the solution is x = 8 + 2 / 293 or 8-2 / 293

Elementary three 1-variable quadratic equation
A warship sails from west to east at the speed of 20 knots, while an electronic reconnaissance ship sails from South at the speed of 30 knots
When the warship reaches point a, the electronic reconnaissance ship is located at point B due south of point a, and ab = 90 nautical miles. If the warship and the reconnaissance ship continue to sail in the original direction at the original speed, Can the reconnaissance ship observe the warship? If so, when can it be observed at the earliest? If not, explain the reason!
HELP SOS

After X hours, the warship and the electronic reconnaissance ship travel to C and D respectively
(Note: C, D position: D on line AB, AC vertical AB, C on the right side of a), then
AD=90-30x,AC=20x
So the distance between two ships CD = root ((90-30x) square + (20x) square)
=Root (8100-5400x + 1300X Square)
When the root number (8100-5400x + 1300X Square) is less than or equal to 50, the reconnaissance ship can detect the warship, then 8100-5400x + 1300X square is less than or equal to 2500
What do you know
That's it
With the original title
13X square - 54x + 56 ≤ 0
2≤x≤28/13
That is, after 2 hours to 28 / 13 hours, the reconnaissance ship can detect the warship

Ask 20 pursuit and itinerary questions in the first volume of seventh grade mathematics. If you want to have difficulty, you'd better attach answers, 10 easy questions for each (use equation solution)

http://wenku.baidu.com/view/988c8 The speed ratio of a and B is 4:5. After the first meeting, the speed of a is increased by one fourth and that of B is increased by one third. When the two cars arrive at Ba, they return immediately. In this way, the second meeting point is 48km away from the first meeting point. How many kilometers is the distance between AB and a?
Consider the total distance as unit 1
Because time is the same, distance ratio is speed ratio
So when we met, a made a total of 1x4 / (5 + 4) = 4 / 9
B: 1-4 / 9 = 5 / 9
At this time, the speed ratio of a and B increases from 4:5 to 4 (1 + 1 / 4): 5 (1 + 1 / 3) = 5:10 / 3 = 3:4
The distance between Party A and Party B is twice the AB distance, which is 2
At this time, the second meeting, B line of the whole 2x4 / (3 + 4) = 8 / 7
The distance of the second meeting point accounts for 8 / 7-4 / 9 of the total distance = 44 / 63
44 / 63-4 / 9 = 16 / 63 from the first meeting point
AB distance = 48 / (16 / 63) = 189km
After school, Xiao Ming walks home at the speed of 4 kilometers per hour along a bus route. Every 6 minutes, a car overtakes him from behind, and every 4 and 2 / 7 minutes, another car comes to him. If the bus keeps running at the same speed at the same time interval, what is the time interval for the issuance of namo bus?
This problem is two processes
The first is the pursuit problem, and the second is the encounter problem
Let the bus speed be a km / h
（a-4）x6/60=（a+4）x(30/7)/60
7a-28=5a+20
2a=48
A = 24 km / h
Then the distance difference between the bus and Xiaoming = (24-4) X6 = 120 km
So the departure interval is 120 / 24 = 5 minutes
The passenger car and the freight car leave from a and B at the same time, and meet after 4 or 5 hours. When they meet, the passenger car travels 27 kilometers more than the freight car, and the speed of the freight car is 4 / 5 of that of the passenger car. How many kilometers are there between B?
The speed ratio of passenger cars to freight cars is 5:4
Think of the whole process as unit 1
That is to say, when the bus meets, it takes 1 × 5 / (4 + 5) = 5 / 9 of the whole journey
Then the freight car is 1-5 / 9 = 4 / 9
Passenger cars travel more than trains 5 / 9-4 / 9 = 1 / 9
So the whole journey = 27 / (1 / 9) = 243 km
This is my own representative, you can refer to c4a767f5acfa1c7cd9a.html

Additional questions: it is known that the speed of a is 20 m / min, and that of B is 25 m / min. they set out from a and B at 8 a.m., first facing each other for 1 minute, then facing back for 3 minutes, then facing each other for 5 minutes, and then facing back for 7 minutes If a and B are 360 meters apart, the first time they meet is ()
A. 9:30 B. 9:20 C. 9:15 D. 9:10

For the first time, walk in opposite direction once, walk in opposite direction once, and walk in opposite direction once again, so as to get close to each other for 3 (20 + 25) meters, and then walk in opposite direction once, so as to get close to each other for 2 (20 + 25) meters. For the following, walk in opposite direction once, so as to get close to each other for 2 (20 + 25) meters, let's start from the seventh minute, and then need n times to get close to each other

A drives from a to a at a speed of 12 km / h, and B pursues a from a at a speed of 16 km / h after 20 minutes. As a result, B overtakes a at a distance of 5 km from B, and finds the distance between a and B

The distance between the two places is x km
(x-5)/12=20/60+(x-5)/16
(x-5)*(1/12-1/16)=1/3
x-5=16
X = 21 km

One variable linear equation of the people's Education Press

The essence of travel problem is the relationship among distance, speed and time, that is, distance = speed × time
As long as you know any two of them, you can work out the third quantity. That is to say, according to this formula, as well as the relationship between speed and speed, the relationship between distance and speed, time and so on, let one of them be x, and the other quantity can be listed according to the relationship between quantity and quantity given in the question