Given the function y = f (x) (x is not equal to 0), for any non-zero real number x, y satisfies f (XY) = f (x) f (y) F (1) = 0, and f (x) is even function If y = f (x) is an increasing function from 0 to positive infinity, the solution to the inequality f (one sixth) + F (X-5) is less than or equal to 0 Teacher's solution is that x belongs to the closed interval from - 1 to 1, and X is not equal to 5. I think the solution is wrong, at least x should be greater than 5

Given the function y = f (x) (x is not equal to 0), for any non-zero real number x, y satisfies f (XY) = f (x) f (y) F (1) = 0, and f (x) is even function If y = f (x) is an increasing function from 0 to positive infinity, the solution to the inequality f (one sixth) + F (X-5) is less than or equal to 0 Teacher's solution is that x belongs to the closed interval from - 1 to 1, and X is not equal to 5. I think the solution is wrong, at least x should be greater than 5

A: F (XY) = f (x) + F (y), right?
F (x) is an even function: F (- x) = f (x)
x> F (x) is an increasing function at 0
Then x

For any nonzero real number x y, the function y = f (x) (x is not equal to 0) satisfies f (XY) = f (x) + F (y) to find f (1) and f (- 1)

For the function f (x), f (XY) = f (x) + F (y)
Let x = y = 1, from F (XY) = f (x) + F (y)
That is, f (1) = f (1) + F (1) = 2F (1)
That is, f (1) = 0
Let x = y = - 1
Let f (1) = f (- 1) + F (- 1) = 2F (- 1)
There is f (- 1) = 0
That is, f (1) = f (- 1) = 0

In the ellipse x square / 4 + y square / 3 = 1, there is a point P (1, - 1), f is the right focus of the ellipse, and there is a point m in the ellipse, which is the minimum of MP + 2mf. What is the value~

You have to use centrifugal rate
You draw a picture, draw a guide line, and then the distance between the point on the ellipse and the guide line is 2
That is to say, when the point (x, y), y = - 1, the minimum value is obtained, and the point is the right half of the ellipse

It is known that there is a point P (1, - 1) in the ellipse x ^ 2 / 4 + y ^ 2 / 3 = 1, F1 is the right focus of the ellipse, and there is a moving point m on the ellipse, the maximum value of | MP | + | MF1 |

Let me tell you how to solve the problem. I'll work out the coordinates of the right focus of the ellipse. Then I'll work out the distance from the fixed point to the right focus, and then I'll work out the sum of the intersection to the two focuses. Let the distance from the intersection to the right focus be x, so the shortest distance required is the distance from the fixed point to the right focus minus x (plus the sum of the intersection to the two focuses - x}

It is known that the left and right focal points of the ellipse x2a2 + y2b2 = 1 (a > b > 0) are F1 and F2 respectively. If there is a point P on the ellipse such that | Pf1 | = e | PF2 |, then the value range of eccentricity e of the ellipse is______ ．

Let the abscissa of point p be x, ∵ Pf1 | = e | PF2 |, then E (x + A2C) = e · e (a2c-x), ∵ x = C − AE (E + 1) can be obtained from the definition of ellipse, and - a ≤ C − AE (E + 1) ≤ a,