Let f (x) = | 2x-a | + 2a, if the solution set of inequality f (x) ≤ 6 is {x | - 6 ≤ x ≤ 4}, find the value of real number a If the solution set of inequality f (x) ≤ (k ^ 2-1) X-5 is not empty, the value range of real number k is obtained

Let f (x) = | 2x-a | + 2a, if the solution set of inequality f (x) ≤ 6 is {x | - 6 ≤ x ≤ 4}, find the value of real number a If the solution set of inequality f (x) ≤ (k ^ 2-1) X-5 is not empty, the value range of real number k is obtained

∵f(x)≤6
∴|2x-a|+2a≤6
|2x-a|≤6-2a
2a-6≤2x-a≤6-a
3a-6/2≤x≤6-a/2
The solution set of ∵ f (x) ≤ 6 is {x | - 6 ≤ x ≤ 4}
∴3a-6/2=-6
6-a/2=4
The solution is a = - 2

Y(2Y+7)=4 ..

2y^2+7y-4=0
(2y-1)(y+4)=0
So y = 1 / 2 or - 4

For the equation of X, if X-8 / X-7 - K / 7-x = 8 has an increasing root, what is the value of K?
If the equation X-1 / X-2 = m / X-2 + 2 of X has no solution, what is the value of M?
Write specific ideas

1. The incrementing root can only be produced because the denominator is 0, that is, the incrementing root x is 7
The equation is X-8 + k = 8 (X-7)
Substituting x = 7 into the solution gives k = 1
2. If the equation has no solution, it can only make x = 2
x-1/x-2=m+2x-4/x-2
Substituting x = 2
1/0=m+4-4/x-2
1/0=m/0
So m = 1

Use a cup to pour water into an empty bottle. If you pour three cups of water into the bottle, the total weight is 440G. If you pour five cups of water into the bottle, the total weight of the bottle is 600g. How much does a cup of water weigh?

A: a glass of water weighs 80g

It is known that f (x) is a decreasing function on (0, + infinity). Try to compare the magnitude of F (square of a minus a plus 1) and f (3 / 4)

The square of a minus a plus 1 = (A-1 / 2) ^ 2 + 3 / 4 is greater than or equal to 3 / 4
F (x) is a decreasing function on (0, + infinity)
So f (square of a minus a plus 1) is less than or equal to f (3 / 4)

The function f (x) defined on R, for any x, y belonging to R, has f (x + y) + F (X-Y) = 2F (x) * f (y), and f (0) ≠ 0
1. Find the value of F (0)
3. If there is a constant C such that f (C / 2) = 0, we prove that f (x + 2C) = f (x) for any x belonging to R

Let y = C / 2, then f (x) = f (x + C / 2) + F (x-C / 2) = 2F (x) f (C / 2) = 0, then f (x + C / 2) = - f (x-C / 2), that is, f (x) = - f (x + C) then f (x + C) = - f (x + C) then f (x + C) = - f (x + 2C), then f (x) = f (x + 2C) is obtained

Mathematical problems about set and function
For the function f (x) defined on R, the following statements are correct: 1 if f (- 2) = f (2), then f (x) is even; 2 if f (- 2) is not equal to f (2), then f (x) is not even; 3 if f (- 2) = f (2), then f (x) is not odd; 4 if f (x) = 0, then f (x) is odd, The following statement is incorrect: the function of a image which is centrosymmetric with respect to the origin must be an odd function; the image of an odd function must pass through the origin; the image of an even function must pass through the origin; if it does not pass through the origin, the number of intersections between it and the X axis must be an even number; the function of D image which is symmetrical with respect to the Y axis must be an even function

(1) The correct definition is 2,4. If you look at the definition more, 4 is a special condition. When you draw a graph, you can see that it is both an odd function and an even function. Later, you can remember that the function image is X-axis (that is, f (x) = 0), which is both an odd and even function
(2) Incorrect is that the first term of B function is to consider the domain problem. Ya Ya Ya, odd function can be defined without x = 0

(sets and functions)
Given the set M = {(x, y) / y = x + K}, n = {(x, y) / x ^ 2 + y ^ 2 = 9}, and M intersection n = empty set, find the value range of real number K
All I know is that a straight line doesn't intersect a circle,
The formula for K seems to have been forgotten~

Find the value of K when the line is tangent to a circle
The distance from the center of a circle to a straight line is equal to the radius
K = plus or minus three root sign two
Because m intersection n = empty set
So K is greater than triple sign two or K is less than minus triple sign two

If f (x) satisfies f (P + Q) = f (P) f (q), f (1) = 3, then f 2 (1) + F 2 (1) + F 2 (2) + F 4 (3) + F 2 (3) + F 6 (5) + F 2 (4) + F 8 (7)=______ ．

From F (P + Q) = f (P) f (q), let P = q = n, F2 (n) = f (2n). The original formula = 2F2 (1) f (1) + 2F (4) f (3) + 2F (6) f (5) + 2F (8) f (7) = 2F (1) + 2F (1) f (3) f (3) + 2F (1) f (5) f (5) + 2F (1) f (7) f (7) = 8F (1) = 24

It is proved that the function f (x) = the square of X + X is an increasing function on (negative half, positive infinity),
Such as the title

Let X1 and X2 belong to (- 1 / 2, positive infinity)
And X1 < x2
From F (x1) - f (x2)
=（x1²+x1）-（x2²+x2）
=（x1²-x2²）+（x1-x2）
=（x1-x2）（x1+x2）+（x1-x2）
=（x1-x2）（x1+x2+1）
From - 1 / 2 < X1 < x2
That is, x1-x2 < 0,
There are also X1 + 1 / 2 > 0, X2 + 1 / 2 > 0, namely X1 + x2 + 1 > 0
That is, (x1-x2) (x1 + x2 + 1) < 0
That is, f (x1) - f (x2) < 0
That is, f (x1) < f (x2)
That is, f (x) = the square of X + X is an increasing function on (negative half, positive infinity)