The function f (x) defined on [0,1] satisfies that f (0) = 0, for any x ∈ [0,1], f (x) + F (1-x) = 1, f (1 / 5x) = 1 / 2F (x), and if 0 ≤ x1

The function f (x) defined on [0,1] satisfies that f (0) = 0, for any x ∈ [0,1], f (x) + F (1-x) = 1, f (1 / 5x) = 1 / 2F (x), and if 0 ≤ x1

F (x) f (1-x) = 1, take x = 1, so f (1) f (0) = 1, so f (1) = 1. The second equation takes x = 1, so f (1 / 5) = 1 / 2F (1), so f (1 / 5) = 1 / 2
Given that O is the origin of coordinates, a (2, - 1), B (- 4,8) and the vector AB + 3, the vector BC = zero vector, the coordinates of the vector OC are obtained
Let C (x, y)
AB vector = (- 6,9)
AB+3BC=0
(-6,9)+3(x+4,y-8)=(0,0）
therefore
3(x+4)=6 x=-6
3(y-8)=-9 y=15
So (- 6,15) is the C point
suppose:C (x,y)
then:(-6,9)+3(x+4,y-8)=(0,0)
so:x=-2 and y=5
so:OC(-2,5)
What does the discriminant of the root of quadratic equation of one variable mean?
The discriminant of the root of quadratic equation with one variable is △ = B ^ 2-4ac
a. B and C are the coefficients of quadratic term, coefficient of linear term and constant term in quadratic equation of one variable
If △ > 0, the equation has two different real solutions, if △ = 0, the equation has two equal real solutions
One variable linear equation 2x-7 (X-2) = - 1 3 (X-2) x 1 = 2x 5 (X-2) + 1 = x - (2x)
Linear equation of one variable
2x-7(x-2)=—1
3 (X-2) x 1 = 2x
5(x—2)+1=x-（2x—1）
17(2—3y)—5(12—y)=8(1—7y)
1.2x-7 (X-2) = - 12x-7x + 14 = - 1-5x = - 15x = 32.3 (X-2) x 1 = 2x3x-6 + 1 = 2XX = 53.5 (X-2) + 1 = x - (2x-1) 5x-10 + 1 = x-2x + 16x = 10x = 5 / 34.17 (2-3y) - 5 (12-y) = 8 (1-7y) 34-51y-60 + 5Y = 8-56y10y = 34y = 3.4 your praise is my driving force
Factorization of a-178; - b-178; - 2A + 2B
a²-b²-2a+2b
=(a+b)(a-b) -2(a-b)
=(a+b-2)(a-b)
a²-b²-2a+2b
=（a-b）（a+b）-2（a-b）
=（a-b）（a+b-2）
(a-b)(a+b-2)
The function defined on R satisfies that f (0) = 0, f (x) + F (1-x) = 1, f (x / 3) = 1 / 2F (x), and
It should be f (x1)
Given points a (- 1,1), B (- 4,5) and vector BC = 3, vector Ba, vector ad = 3, vector AB, vector AE = half vector AB, find the coordinates of points c, D, e
BC=3BA, AD=3AB, AE=(1/2)AB
OC
= OB +BC
= OB + 3BA
= OB + 3(OA -OB)
=3OA-2OB
=(-3,3) -(-8,10)
=(5,-7)
C (5,-7)
OD = OA +AD
= OA + 3AB
= OA+3(OB-OA)
= 3OB - 2OA
= (-12,15)-(-2,2)
= (-10,13)
D(-10,13)
OE = OA+AE
=OA +1/2AB
= OA +1/2(OB-OA)
= 1/2(OB+OA)
=1/2((-4,5)+(-1,1))
= (-5/2, 3)
E(-5/2,3)
Set point C = (x1, Y1)
BC=3BA ===>(x1+4,y1-1)=3(3,-4)
That is, X1 + 4 = 9; y1-1 = - 12
The solution is X1 = 5; Y1 = - 11
Let d = (X3, Y3)
Ad = 3AB, i.e. (X3 + 1, y3-1) = 3 (- 3,4)
That is, X3 + 1 = - 9; y3-1 = 12
The solution is X3 = - 10; Y3 = 13
Let e = (X2, Y2)
AE=AB/2 ===>(x2+1,y2-1)=(-3,4)/2
That is, X2 + 1 = - 3 / 2; y2-1 = 2
The solution is x2 = - 5 / 2; y2 = 3
C（5，-7）、D（-10，13）、E（-5/2，3）
Discriminant of the root of quadratic equation with one variable
If the quadratic equation (ab-2b) x ^ 2 + 2 (B-A) x + 2A AB = 0 has two equal real roots, then 1 / A + 1 / b =?
The quadratic equation (ab-2b) x ^ 2 + 2 (B-A) x + 2A AB = 0 has two equal real roots
So: △ = 0
So: [2 (B-A)] ^ 2-4 (ab-2b) * (2a AB) = 0
Simplification: 4 (a-b) ^ 2-4ab (2a-ab-4 + 2b) = 0
Expand and simplify: A ^ 2 + B ^ 2 + A ^ 2B ^ 2-2a ^ 2b-2ab ^ 2 + 2Ab = 0
Merge: (a + b) ^ 2 + (AB) ^ 2-2ab (a + b) = 0
Let m = a + B, n = ab
The original formula becomes: m ^ 2-2mn + n ^ 2 = 0
Divide both sides by n ^ 2
We get: (M / N) ^ 2-2 (M / N) + 1 = 0
Merge: [(M / N) - 1] ^ 2 = 0
Square root on both sides: M / n-1 = 0
Transfer: M / N = 1
Because: M = a + B, n = ab
So: a + B / AB = 1
That is: 1 / A + 1 / b = 1
The result is 1
The method is as follows: the equation has two equal solutions satisfying the formula (bsquare - 4AB = 0). After simplification, it is 4A square + 8ab + 4B square - 8ab (a + b) + 4 (AB) square = 0
(2a + 2b-2ab) square = 0
So a + B = ab
[delta]=[2(b-a)]^2-4*(ab-2b)*(2a-ab)=0
If the quadratic equation AX ^ 2 + BX + C = 0
Then [delta] = B ^ 2-4ac
If the original equation has two unequal real roots, then [delta] > 0
If the original equation has two equal real roots, then [delta] = 0
If the original equation has no real root, then [delta]
3 (x + 2) - 5 = 4 (2x-6) univariate linear equation
4(2x+3)=8-3(x+1)
3(x+2)-5=4(2x-6)
3x+6-5=8x-24
8x-3x=1+24
5x=25
X=5
4(2x+3)=8-3(x+1)
8x+12=8-3x-3
8x+3x=5-12
11x=-7
x=-7/11
Factorization (2a-b) & #178; - (a-2b) & #178;