Whose book is that

Whose book is that

Whose book is that?
Upstairs is "whose book is that"

Don't put the book here

Don't put the book here.

Do you need sports bags

Do you need to exercise bag?

Mathematical Olympiad: 53.5 times 35.3 plus 53.5 times 43.2 plus 78.5 times 46.5

53.5*35.3+53.5*43.2+78.5*46.5
=53.5*(35.3+43.2)+78.5*46.5
=53.5*78.5+78.5*46.5
=78.5*(53.5+46.5)
=78.5*100
=7850

RT. find the special solution of differential equation: y '' + (y ') ^ 2 = 1, when x = 0, y = y' = 0

Let P = y ', p * DP / dy + P ^ 2 = 1
The corresponding homogeneous equation is p * DP / dy = - P ^ 2
dp/p=-dy
ln|p|=-y＋ln|C|
P = CE ^ (- y)
By the method of constant variation, P = UE ^ (- y)
Substituting p * DP / dy + P ^ 2 = 1, the solution is UDU / dy = e ^ (2Y)
That is u ^ 2 / 2 = 1 / 2 * e ^ (2Y) + C '/ 2
u=√(e^(2y)＋C1)
That is, P = e ^ (- y) √ (e ^ 2Y + C1)
And y = P = 0, C1 = - 1
dy/dx=√(1-e^(-2y)
So dy / √ (1-e ^ (- 2Y)) = DX
-ln[(1-√(1-e^(-2y))/e^(-y)]=x-ln|C2|
Substituting x = y = 0, we get ln | C2 | = 0
So (1 - √ (1-e ^ (- 2Y)) / e ^ (- y) = e ^ (- x)
So the special solution of the differential equation is 1-e ^ (- x) e ^ (- y) = √ (1-e ^ (- 2Y))

The equation x ^ 2-ax + 4 = 0 of X has a solution on [- 1.1]. Find the value range of real number a

X = 0 is obviously not the solution,
So there is a = (x ^ 2 + 4) / x = x + 4 / X
|X + 4 / X | > = 2 √ (x * 4 / x) = 4, when | x | = 2, take the equal sign, in | x | = | 1 + 4 | = 5
So the value range of a is: a > = 5 or a

45 / 7, 2 column vertical calculation

The definition of higher mathematics limit
The limit of function has nothing to do with whether f (x) is defined at point x0

That is to say, the function is not defined at this point, or the domain does not contain this point
Let's take an example
F (x) = x + 1, the definition field is that x is not equal to 1
Obviously, the function is undefined when x = 1, but the limit at x = 1 exists

In the bivariate linear equation 4x + 3Y = 7, if x and y are opposite to each other, then x = [], y = []. How to calculate its answer

In the quadratic equation 4x + 3Y = 7, if x and y are opposite numbers, then x = [3], y = [- 3]
solution
If x and y are opposite to each other, then
x+y=0 （1）
And 4x + 3Y = 7 (2)
(1) * 4 - (2) de
y=-7
Substituting (1) gives x = 7

How to make 123 = 11234 = 112345 = 1

(1 + 2) divide by 3 = 1,1 times (2 + 3) - 4 = 1,1-2 + 3 + 4-5 = 1