Which of the following questions can be answered Whoever answers more and better will get 100 points, The first question is the fractional equation, and the second and third questions are the solutions 1、 (25/x)+1=(30/2x) 2、 C is the midpoint of a segment AB and D is a point on CB. As shown in Figure 3, if the lengths of all segments are positive integers and the product of all possible lengths of AB is equal to 140, what is the sum of all possible lengths of AB? 3、 For the integer 6x ^ 5 + 5x ^ 4 + 4x ^ 3 + 3x ^ 2 + 2x + 2002, given a value, if Xiao Ming calculates the value of the integer according to the rules of four operations, he needs to calculate 15 times of multiplication and 5 times of addition. Xiao Hong said: there is another algorithm, as long as the appropriate brackets are added, the number of addition can not be changed, while the number of multiplication is only 5 times Is Xiaohong right or wrong? If it is right, please explain the reason: what does Xiaohong think This is the picture of question 2 A------C----D--B The third question is 6 (x ^ 5) + 5 (x ^ 4) + 4 (x ^ 3) + 3 (x ^ 2) + 2x + 2002 Don't get it wrong

Which of the following questions can be answered Whoever answers more and better will get 100 points, The first question is the fractional equation, and the second and third questions are the solutions 1、 (25/x)+1=(30/2x) 2、 C is the midpoint of a segment AB and D is a point on CB. As shown in Figure 3, if the lengths of all segments are positive integers and the product of all possible lengths of AB is equal to 140, what is the sum of all possible lengths of AB? 3、 For the integer 6x ^ 5 + 5x ^ 4 + 4x ^ 3 + 3x ^ 2 + 2x + 2002, given a value, if Xiao Ming calculates the value of the integer according to the rules of four operations, he needs to calculate 15 times of multiplication and 5 times of addition. Xiao Hong said: there is another algorithm, as long as the appropriate brackets are added, the number of addition can not be changed, while the number of multiplication is only 5 times Is Xiaohong right or wrong? If it is right, please explain the reason: what does Xiaohong think This is the picture of question 2 A------C----D--B The third question is 6 (x ^ 5) + 5 (x ^ 4) + 4 (x ^ 3) + 3 (x ^ 2) + 2x + 2002 Don't get it wrong

1. X = - 102. AB can be divided into two equal parts, and they are all integers, so AB is even. CB can also be divided into two integer long segments, so CB > = 2. So AB must be an even number of > = 4. 140 = 2 * 2 * 5 * 7 can only be divided into two even sums greater than 2 * 5 and 2 * 7 (it must be divided into the multiplication product of two or more numbers

Three math problems 100 points urgent
① If a.b.c.x.y.z. ∈ R, and satisfy B & sup2; > 0 of AC-B, AZ + 2by + CX = 0. XYZ ≠ 0, prove: xz-y & sup2; = 0
② X.y.z ∈ R, prove X & sup2; - XZ + Z & sup2; + 3Y (x + Y-Z) ≥ 0
③ Let A.B.C be positive numbers, and the following inequalities are proved
1. At least one of a + B < C + D 2. (a + b) (c + D) < AB + CD 3 (a + b) CD < AB (c + D) is incorrect
Finally, it's better to use the counter evidence (it's OK not to use it) and if you have a wrong number, please solve the other problems as soon as possible

The idea of the second problem is as follows: write the left side in the form of quadratic equation with one variable of X
X ^ 2 + (3y-z) x + (Z ^ 2 + 3Y ^ 2-3yz) = 0
(3y-z)^2-4(z^2+3y^2-3yz)=-3z^2-3y^2+6yz=-3(y-z)^2

Three Mathematical Olympiad questions, still 100 points
1. The apprentice and the apprentice process 12 parts per hour. The apprentice has processed 8 hours before the master starts to process. The master processes 18 parts per hour. After a few hours, the apprentice and the apprentice process the same parts
2. Xiao Wang and Xiao Li have a meeting from place a to place B. Xiao Wang walks 15 kilometers per hour on his bicycle. Xiao Li sets out on a motorcycle only when he is 2 hours away. It takes Xiao Li an hour to catch up with Xiao Wang and ask him how many meters per hour on his motorcycle?
3. The PLA carries out the marching task. The troops start from a certain place and travel 112 kilometers per hour. When they are 6 hours old, the correspondent rides a motorcycle to chase the troops at a speed of 48 kilometers per hour to convey the order. How many hours can he catch up with the troops?
Wrong number 3. The troops drive 12 kilometers per hour.

1. Time = 12 × 8 ÷ (18-12) = 16 hours
After 16 hours, the parts processed by master and apprentice are equal
2. Speed = 15 × (2 + 1) △ 1 = 45 km / h
Xiao Li rides a motorcycle at 45 kilometers an hour
3. Catch up = 12 × 6 △ (48-12) = 2 hours
We can catch up with the troops in two hours

100 points
1. The side length of a square is 3cm. When the side length of the square is reduced by xcm, the perimeter of the new square is YCM. Then the functional relationship between X and Y is
2. Given that the perimeter of a rectangle is 12 and the length of one side is x, then the value range of the functional relationship between its area y and X Wei x is?
3. When x=____ The value of y = [X-1] of [X-1] is 0?
4. In the equation 2x-3y = 5, if y = - 1 and X is not equal to 0, D: x > = 1 and X is not equal to 0
6. Given the function y = [2x-1] of [x + 2], when x = a, the function value is 1, then the value of a is:
A.1 B.3 C.-3 D.-1

1.y=-4x+12
two

Given that two of the square of the equation x - (2t + 3) x + 2T = 0 satisfy that x 1 is less than 2 and x 2, the value range of T is obtained

Let f (x) = x ^ 2 - (2t + 3) x + 2T
Two satisfy x1

It is known that the equation x ^ 2 - (T-2) x + T ^ 2 + 3T + 5 = 0 has two real roots, a vector = (- 1,1,3), B vector = (1,0, - 2), C vector = a + t
When | C | is the minimum, the value of T is obtained
Modify C vector = a + TB

According to the fact that the equation has two real roots, △ > = 0, the range of t can be calculated [- 4, - 4 / 3]
c=a+tb=(-1+t,1,3-2t)
|c|=√5t^2-14t+11
According to the image of quadratic function, we know that y = 5T ^ 2-14t + 11, the vertex coordinate (7 / 5,6 / 5) function decreases on (- ∞, 7 / 5], so when | C | takes the minimum value, t = - 4 / 3

If the square of equation x + 2mx + 2 = 0 has negative root, then the value range of real number m is?

First, the discriminant is greater than or equal to 0
4m²-8>=0
m=√2
x=[-2m±√(4m²-8)]/2
Obviously, the root of a minus sign is smaller
Then as long as [- 2m - √ (4m & sup2; - 8)] / 2 = √ 2, then - M0
square
m²-2>m²
-2> 0, does not hold
So m > = √ 2

If the quadratic function f (x) satisfies f (x + 4) + F (x-1) = x & # 178; - 2x, then f (x)=

Let f (x) = ax & # 178; + BX + C
Then, f (x + 4) + F (x-1) = 2aX & # 178; + (2b-6a) x + (17a + 3B + 2C) = x & # 178; - 2x
So,
2a＝1
2b－6a＝－2
17a＋3b＋2c＝0
Solution
a=1/2,b=1/2,c=-5
So,
f（x）= x²/2＋x/2 -5

Let f (x) = ax ^ 2 + BX + C
Is it based on the fact that f (x) is a quadratic function?
The later f (x + 1) - f (x) = a (x + 1) ^ 2 + B (x + 1) + c-ax ^ 2-bx-c
What is the reason?
The detailed process of this problem is to find the expression of F (x) when f (x) is known to be a quadratic function, satisfying f (1 + x) + F (1-x) = 2x & # 178; + 6, f (0) = 1

Since f (x) is a quadratic function in the title, the general expression of F (x) can be set
f(x)=ax^2+bx+c
If f (0) = 1, then C = 1
f(1+x)=a(1+x)^2+b(1+x)+1
f(1-x)=a(1-x)^2+b(1-x)+1
And f (1 + x) + F (1-x) = 2x & # 178; + 6;
We get 2aX ^ 2 + 2 (a + B + 1) = 2x ^ 2 + 6
Compare the coefficients
We get a = 1, B = 1
So f (x) = x ^ 2 + X + 1

Given the quadratic function f (x + 1) = f (1-x) and f (0) = 0, f (1) = 0, if f (x) is a closed interval from m to N on the closed interval from m to 0, find the value of M, n

M to 0? M to n? And F (x + 1) = f (1-x) here, we can see that the axis of symmetry is x = 1, but f (0) = 0, f (1) = 0, where the axis of symmetry becomes x = 1 / 2, which is contradictory to = = the correct problem should be as follows: it is known that the quadratic function f (x) satisfies f (x + 1) = f (1-x), and f (0) = 0, f (1) = 1