Please use vedadine to understand the equation （k2+1）x2+2k（1-k）x+（1-k）2-2=0

Please use vedadine to understand the equation （k2+1）x2+2k（1-k）x+（1-k）2-2=0

The factorization factor of cross phase multiplication is [(k ^ 2 + 1) x - (k ^ 2 + 1)] [x - (k ^ 2-2k-1) / (k ^ 2 + 1)] = 0
So the solution of the equation is 1, and (k ^ 2-2k-1) / (k ^ 2 + 1)
The odd function f [x] defined on 〔 - 1,1 〕 is a decreasing function, which solves the inequality about a; f [1-A] + the square of F [1-A] is less than 0
f(1-a)+f(1-a^2)
What grade are you in? ,
1. Let u = R set a = {X / X ＜ - 1} set B = {X / - 2 ≤ x ＜ 3} find CUA ∪ cub CUA ∩ cub
2. Let u = {X / - 3 ≤ x ≤ 5} a = {X / - 1 ＜ x ≤ 1} B = {X / 0 ≤ x ＜ 2} find a intersection B, a union B, CUA, Cub CUA ∪ cub CUA ∩ cub
1. CUA = {x | x ≥ - 1}, Cub = {x | x < - 2 or X ≥ 3}, so
CUA ∪ cub = {x | x < - 2 or X ≥ - 1},
CuA∩CuB=｛x | x≥3｝.
2、A∩B=｛x | 0≤x≤1｝,A∪B=｛x | -1
How to prove Veda theorem
If the odd function f (x) defined on (- 1,1) is a decreasing function and f (1-A) + F (1-A 2) < 0, the value range of real number a is obtained
∵ f (x) is an odd function ∵ f (1-A) < - f (1-a2) = f (A2-1) ∵ f (x) is a decreasing function defined on (- 1, 1) ∵ 1 ∵ a ∵ 1 ∵ a ∵ 1 ∵ a ∵ 1 ∵ a ∵ 1 ∵ a ∵ a ∵ 2 ∵ 1 ∵ a ∵ a ∵ 1 ∵ a
Let u = {x | - 3 be greater than or equal to X and less than or equal to 5}, a = {x | - 1}
Set u = [- 3,5], a = (- 1,1), B = [0,2), CUA = [- 3, - 1] U (1,5] cub = [- 3,0) u [2,5], and then write the form of set ～
How to prove Veda theorem
prove:
When Δ = B ^ 2-4ac ≥ 0, the equation
ax^2+bx+c=0(a≠0)
There are two real roots, set as x1, x2
From the root formula x = (- B ± √Δ) / 2a, we can take
x1＝(-b-√Δ)/2a,x2＝(-b+√Δ)/2a,
Then: X1 + x2
=(-b-√Δ)/2a+(-b+√Δ)/2a
=-2b/2a
=-b/a,
x1*x2=[(-b-√Δ)/2a][(-b+√Δ)/2a]
=[(-b)^2-Δ]/4a^2
=4ac/4a^2
=c/a.
In conclusion, X1 + x2 = - B / A, X1 * x2 = C / A
Given that f (x) is an increasing function defined on (0, + ∝), f (XY) = f (x) + F (y), f (2) = 1, find the solution set of the inequality f (x) + F (X-2) > 3
..
f(8)=f(2)+f(4)=f(2)+f(2)+f(2)=3f(2)=3
We know that f (x) is defined on (0, + ∝), so x > 0, X-2 > 0
It is known that f (x) is an increasing function defined on (0, + ∝)
Therefore, f (x) + F (X-2) = f (x ^ 2-2x) > 3 = f (8) to get: x ^ 2-2x > 8
The solution is: x > 4, or X4
f（2）=f（2^1）=1，f(4)=f(2^2)=2,。。。 We can get that f (x) = log2 (x), (2 is subscript), f (x) + F (X-2) = log2x + log2 (X-2), log2x + log2 (X-2) > 3 can get x > 4 or X4
Let u = {x | 0
So {a, B = {1,457},
And Cu (a ∪ b) = (CUA) ∩ (cub) = {9}, so a ∪ B = {1,2,3,4,5,6,7,8},
So B = {2,3,4,6,8}
It's very convenient to do this kind of drawing Venn diagram
Because u = {x ∈ n | x ≤ 8}, u = {1,2,3,4,5,6,7,8}. Because a ∩ (cub) = {2,8}, there are 2,8 in a. There is no 2,8 (CUA) ∪ (cub) = Cu (a ∩) in B
What does Veda's theorem mean? It should be specified
There is a certain relationship between the two roots of a quadratic equation of one variable and the coefficients of the equation. The mathematician Weida discovered and proved that the sum of two roots of a quadratic equation of one variable equals to the coefficient of negative first order divided by the coefficient of second order: X1 + x2 = - B / A. the product of two roots equals to the coefficient of constant divided by the coefficient of second order: X1 * x2 = C / A
The relationship between equation and coefficient: x1x2 = C / A, x1x2 = - B / A
Weida theorem proves the relationship between root and coefficient in the equation of degree n with one variable.
Give an example to illustrate the relationship between two quadratic equations of one variable
The quadratic equation AX ^ 2 + BX + C = 0 (a is not equal to 0),
The two x1, X2 of the equation and the coefficients a, B, C of the equation satisfy X1 + x2 = - (B / a), X1 * x2 = C / A.