④ Known: (x-2013) (2014-x) = m, find the value of (x-2013) 2 + (2014-x) 2 [expressed by the algebraic formula containing M]

④ Known: (x-2013) (2014-x) = m, find the value of (x-2013) 2 + (2014-x) 2 [expressed by the algebraic formula containing M]

(x-2013)^2+(2014-x)^2
=[(x-2013)+(2014-x)]^2-2*(x-2013)(2014-x)
=1-2m
2 outside the brackets in the title should be a square, right

Given that the solution of the equation x + 1 / x = m + 1 / m about X is X1 = m, X2 = 1 / m, what is the solution of the equation x + 1 / (x-1) = m + 1 / (m-1) about x

X+1/（X-1）=m+1/(m-1)
So X-1 + 1 / (x-1) = M-1 + 1 / (m-1)
So x1-1 = M-1
x2-1 = 1/(m-1)
So X1 = M
x2 = 1/(m-1)+1

On the two real roots X1 and X2 of the equation x ^ (m ^ + 1) x + m-2 = 0, which satisfies the condition (1 + x1) (1 + x2) = 2,4, we can find the value of M

The title is this
It is known that two real numbers of the equation x & sup2; - (M & sup2; + 1) x + m-2 = 0 with X1 and X2 satisfy (1 + x1) (1 + x2) = 2, and the value of M is obtained
（1+X1)(1+X2)=2
1+X2+X1+X1X2 =2
X1+X2+X1X2=1
Weida theorem X1 + x2 = - B / a x1x2 = C / A
-（-M²-1)+M-2=1
M²+1+M=1+2
M²+M-2=0
Cross multiplication (M + 2) (m-1) = 0
‖ m + 2 = 0 or M-1 = 0
That is M1 = - 2, M2 = 1
Hey, hey, I made it myself

Using the convergence criterion of monotone bounded sequence, the existence of the following sequence limit is proved
X1 = radical 2, X (n + 1) = radical 2x, n = 1,2,3

1.x1=√2

Train a and train B run from two places 700 kilometers apart at the same time. Train a runs 85 kilometers per hour and train B runs 90 kilometers per hour. How many hours do the two trains meet?

700 △ 85 + 90 = 700 △ 175, = 4 hours. A: the two trains meet in 4 hours

Given that the average of data a, B, C, D is 4, then the average of a group of new data 2a-3, 2b-3, 2c-3, 2d-3 is ()
Urgent

The average of data a, B, C and D is 4, so a + B + C + D = 4 * 4 = 16
The sum of 2a-3, 2b-3, 2c-3, 2d-3 is 2a-3 + 2b-3 + 2c-3 + 2d-3 = 2 (a + B + C + D) - 3 * 4 = 2 * 4 * 4-3 * 4 = 20
20/4=5
The average of new data 2a-3, 2b-3, 2c-3, 2d-3 is (5)

A. The distance between B and a is 50km. A rides a bicycle from a to B. after one hour and 30 minutes of departure, B rides a motorcycle from a to B
The speed of a is 2.5 times that of B, and B is 1 hour earlier than A. what are the speed of a and B?

If the speed of a is x km / h, the speed of B is 2.5 x km / h
50÷x = 50÷2.5x + 1+ 1.5
Solution
X = 12 km / h
The speed of B is 2.5x = 30 km / h

A natural number greater than 0 divided by 2 is the square of a natural number, and the quotient divided by 3 is the cube of a natural number

Let this number be n, then: n / 2 = a ^ 2n / 3 = B ^ 3, that is 2A ^ 2 = 3B ^ 3, so a is a multiple of 3, B is a multiple of 2, so a = 3A1, B = 2B1, then 2 * 9a1 ^ 2 = 3 * 8b1 ^ 33a1 ^ 2 = 4b1 ^ 3, so A1 = 2A2, B1 = 3B23 * 4a2 ^ 2 = 4 * 27b2 ^ 3a2 ^ 2 = 9b2 ^ 3, so B2 = n ^ 2, A2 = 3N ^ 3, so n = 2A ^ 2 = 2 (3A1) ^ 2 = 18a1 ^ 2 = 1

Two cars a and B are facing each other from a and B at the same time. The speed of car B is 80% of that of car a. the two cars meet at the distance of 30 km from the midpoint. The speed of car a and B is higher than that of car a
What's the distance?

A. B the distance between the two places is:
30*2/(1-80%)*(1+80%)
=60/0.2*1.8
=540km .

In the quadrilateral ABCD, ∠ D + 60 °∠ B is 20 ° larger than ∠ a, and ∠ C is twice as large as ∠ A. find out the size of ∠ a, ∠ B, ∠ C

Let the degree of angle a be X,
Then the angle B is x + 20
The angle c is 2x
The sum of internal angles of quadrilateral is 360
That is x + (x + 20) + 2x + 60 = 360
The solution of the equation is x = 70
Then angle a is 70 degrees, angle B is 90 degrees and angle c is 140 degrees