For the function f (x) whose domain belongs to R, f (a, b) = AF (b) + BF (a), and the absolute value of F (x) is less than or equal to 1 This is a subject of independent enrollment

For the function f (x) whose domain belongs to R, f (a, b) = AF (b) + BF (a), and the absolute value of F (x) is less than or equal to 1 This is a subject of independent enrollment

Let a = b = 0, then f (0) = 0, for any non-zero real number x0, let B = x0, then f (ax0) = AF (x0) + x0f (a), when a ≠ 0, f (x0) / x0 = f (ax0) / ax0-f (a) / A, because LIM (a →∞) (1 / a) = 0, │ f (a) │ ≤ 1, so LIM (a →∞) (f (a) / a) = 0, because LIM (a →∞) (1 / (ax0)) = 0, │ f (ax0) │ ≤ 1

Let f (x) for all x not equal to 0, satisfy AF (x) + BF (1 / x) = C / x, where a, B, C are constants and the absolute value of a is not equal to the absolute value of B, find the expression of F (x)

Because AF (x) + BF (1 / x) = C / X
Let x = 1 / X
Then AF (1 / x) + BF (x) = CX (2)
From ① * B to ② * a
(a²-b²)f(x)=ac/x-bcx
Because the absolute value of a is not equal to the absolute value of B
So a & # 178; - B & # 178; is not equal to 0
f(x)=c(a/x-bx)/(a²-b²)

Given the function f (x) = x + 2 (0b > = 0, and f (a) = f (b), then the value range of BF (a) is obtained

Because both of them are increasing functions, so a > = 1, = 2, so B + 3 / 2 > = 2, so b > = 1 / 2, and 0

Given that a and B are positive real numbers, a + B = 1, X1 and X2 are positive real numbers, it is proved that (ax1 + bx2) (BX1 + AX2) is greater than or equal to x1x2

Expand the left formula to prove the conclusion: abx1 ^ 2 + abx2 ^ 2 + (a ^ 2 + B ^ 2) * x1x2 ≥ x1x2, that is, ab (x1 ^ 1 + x2 ^ 2) + (a ^ 2 + B ^ 2) * x1x2 ≥ x1x2. Because X1 and X2 are positive real numbers, so X1 ^ 1 + x2 ^ 2 ≥ 2x1x2, then the left formula ≥ AB (2x1x2) + (a ^ 2 + B ^ 2) * x1x2 = x1x2 (a ^ 2 + 2Ab + B ^ 2) = x1x2 (a + b) ^ 2 = x1x2 holds

Use 20 cm long and 8 cm wide tiles to lay a square floor. What is the minimum side length of the square floor?
How many centimeters is it?

The least common multiple of 20 8 is 40
So the minimum side length is 40 cm

Kirchhoff's current law can be expressed as?

I in = I out
Condition: for any node in any lumped parameter circuit, at any time, the algebraic sum of current of each branch leaving the node is zero
Have a nice day

The limit of (x-1) arcsinx when x tends to 1

The answer should be 0
To find the limit of (x-1) arcsinx when x tends to 1, its two parts (x-1) and the limit value of arcsinx are solvable. (x-1) when x approaches 1, the extremum is 0, that is, infinitesimal. When x approaches 1, the extremum is half π, that is, finite value. According to the theorem, the finite value of infinitesimal multiplication is still infinitesimal, and the answer is 0;

A large cube is cut into eight equal small cubes. The sum of the surface area of these small cubes is () times larger than that of the original cube
A. 12B. 1C. 2D. 14

It is known that a = 0.0 06, B = 0.0 (2006 zeros in the middle) 025, then a is divided by B=（

2.4. Or 12 / 5

Finding the general solution dy / DX = sin (X-Y)

∵ dy / DX = sin (X-Y) = = > dy = sin (X-Y) DX = = > DX dy = DX sin (X-Y) DX = = > d (X-Y) = (1-sin (X-Y)) DX = = > d (X-Y) / (1-sin (X-Y)) = DX = = > d (X-Y) / (sin ((X-Y) / 2) - cos ((X-Y) / 2)) ^ 2 = DX (application (sin ((X-Y) / 2) ^ 2 + (COS ((X-Y) / 2) ^ 2 = 1) = = > (SEC ((X-Y) / 2