Let f (x) be a function on R and satisfy f (0) = 1, and for any real number x, Y. if f (X-Y) = f (x) - Y (2x-y + 1), the analytic solution of F (x) is obtained

Let f (x) be a function on R and satisfy f (0) = 1, and for any real number x, Y. if f (X-Y) = f (x) - Y (2x-y + 1), the analytic solution of F (x) is obtained

The identity f (X-Y) = f (x) - Y (2x-y + 1) can be reduced to f (X-Y) - (X-Y) & sup2; - (X-Y) = f (x) - X & sup2; - X
Let g (x) = f (x) - X & sup2; - x, then G (X-Y) = g (x) holds for any real number x, y
х g (x) is a constant function,
And G (0) = f (0) = 1,
That is, f (x) = x & sup2; + X + 1

Let f (x) be a function on R and satisfy f (0) = 1, and for any real number x, y, f (X-Y) = f (x) - Y (2x-y + 1), find the analytic expression of F (x)
Let x = 0, then f (0-y) = f (0) - Y (0-y + 1),
That is, f (- y) = 1-y (1-y) = y2-y + 1
Let - y = x, then f (x) = x2 + X + 1
Why let - y = x? Is that something that can be replaced?

Yes, because he is an unknown. It's amazing. If you don't like him, change it to T. it doesn't matter
The meaning of F (x) is that for the unknown x, there is the following rule
Then x is the representation of an unknown number, and - y is also the representation of an unknown number. What's the difference between them?
Change back and forth, it's amazing, but in fact, it's like that

1: The set composed of all real roots of equation X-9 = 0 2: the set composed of all prime numbers of small and 8 3: the set composed of the intersection of the graph of the linear function y = x + 3 and y = --- 2x + 6 4: inequality 4x-5

1. X-9 = 0, x = 9, x = 3 or x = - 3 ｜ the set composed of all the real roots of the equation X-9 = 0 is {3, - 3} 2, the prime numbers less than 8 are 2,3,5,7 ｜ the set composed of all the prime numbers less than 8 is {2,3,5,7} 3, the set composed of solving the equation y = x + 3 ① y = - 2x + 6 ② is x = 1, y = 4 ｜ the intersection coordinate is (1,4) ｜ the set composed of the intersection of the primary function y = x + 3 and the image of y = - 2x + 6 is {(1,4), 4) Pay attention not to leave out 4 and 4x-5 in the middle bracket

Let f (x) be a function on R and satisfy f (0) = 1. For any real number x, y, f (X-Y) = f (x) - Y (2x-y = 1), the analytic expression of F (x) is obtained

Because f (0) = 1
When y = x, f (x-x) = f (x) - x (2x-x + 1) = f (x) - x (x + 1)
That is: F (0) = f (x) - (x ^ 2 + x)
That is, 1 = f (x) - (x ^ 2 + x)
f(x)=-x^2+x+1

The set of real number solutions of the equation x ^ 2 + 2 = 0
Is the real solution of the equation x ^ 2 + 2 = 0 a set

An empty set is also a set

The set of real number solutions of equation x with square + 2 = 0
I'm Xiaobai. Please answer me
The solution set of sum x squared minus 1 = 0

x²+2=0
x²=-2
∵x²≥0
The equation has no real solution

The elements in the set a consist of the solution of the equation KX with respect to x, where k is a real number. If there is only one element in a, the value of K is obtained
Try to have more process so that I can understand

There is only one element in set a, and the equation has equal real roots,
The discriminant is 0,
9-8K=0
k=9/8

Given the set a = {X / ax ^ 2 + X + 1 = 0, X belongs to R}, and a ∩ {X / X ≥ 0} = empty set, find the value range of real number a
Given the set a = {X / ax ^ 2 + X + 1 = 0, X belongs to R}, and a ∩ {X / X ≥ 0} = empty set, find the value range of real number a

That is, X in a has no solution greater than or equal to zero
ax²+x+1=0
Δ=1-4a
x=(-1±√Δ)/2a
So there are three conditions for a
(1) ax²+x+1=0 =>a ≠ 0
(2) Δ = 1-4a = > 1-4a > 0 = > A (- 1 + √ Δ) and (- 1 - √ Δ) must have the same sign, otherwise the two roots are positive and negative
=>Both must be less than 0 (because - 1 - √Δ is already less than 0)
=>Because X has only roots less than 0, and the numerator of x = (- 1 ± √Δ) / 2a is less than 0, the denominator must be positive, that is, 2A > 0
=>a>0
-1 + √ Δ 0, so Δ

Given the set a = {x | X & sup2; + 4x = 0}, B = {x | X & sup2; + 2 (a + 1) x + A & sup2; - 1 = 0, X ∈ r}, a ∩ B = B, find the value of real number a
Finding the value range of real number a

x|x²＋4x＝0
x(x+4)=0
x=0.x=-4
X|x & sup2; + 2 (a + 1) x + A & sup2; - 1 = 0
When x = 0, a = 1, a = - 1
When x = - 4, a = 1, a = 7
A ∩ B = B so - 1 to 7

Suppose a = {x x & sup2; + (P + 2) x + 1 = 0, X ∈ r}, a ∩ positive real number = empty set, find the value range of real number P

P>=0 or P