Let u = R set a = {the square of X / x minus 4x plus 3 is less than or equal to 0}, B = {the square of X / x minus 2x is greater than 0} find (1) a, B (2) (CUA) U (cub)

Let u = R set a = {the square of X / x minus 4x plus 3 is less than or equal to 0}, B = {the square of X / x minus 2x is greater than 0} find (1) a, B (2) (CUA) U (cub)

x²-4x+3≤0
The solution is a = {x | 1 ≤ x ≤ 3}
x²-2x>0
The solution is: B = {x | x > 2 or X3 or X3 or X ≤ 2}
From x ^ 2-4x + 3 ≤ 0
(x-3) (x-1) ≤ 0
The solution is 1 ≤ x ≤ 3
That is, a = {x | 1 ≤ x ≤ 3}
So CUA = {x | X3}
From x ^ 2-2x > 0
X > 2 or x2 or X3}
(1)A={x/1
Veda's theorem for quadratic equation of one variable
The relationship between the two elements of the equation and the numbers in the equation is as follows: X1 + x2 = - B / A, x1 · x2 = C / a (also known as WIDA's theorem)
When the two equations are X1 and X2, the equation is: X2 - (x1 + x2) x + x1x2 = 0 (obtained from the inverse deduction of Weida theorem)
1. Let F X be an odd function and f x-f y = f (X-Y) for any XY belonging to R. if x 0 f (1) = - 5, find the maximum value of F (x) on [- 2,2]
2. It is known that a = {x belongs to R, x2-2x-8 = 0} B = {x ∈ R, X2 + ax + a2-12 = 0} a ∪ B = B
Finding the value range of real number a
3. Let a = {x 2 ≤ x ≤ 7} B {2m ≤ x ≤ 3M + 1} find the value range of real number m, so that CRA ∩ B = empty set
4. Let u = R, a = 0 ≤ X
Let u = R, a = 0 ≤ X
Let u = R, a = {x | - 2 ≤ x ＜ 4}, B = {x | - 8-2x ≥ 3x-7}, find Cu (AUB), CUA ∩ cub
The solution is: B = {x | x ≤ 3},
A∪B={x|x
How to do the root of quadratic equation with one variable and Weida theorem
55. I didn't understand,
ax^2+bx+c=0
（a≠0）
Sum of two = - (B / a)
Product of two = C / A
If B ^ 2-4ac > 0, the equation has two unequal real roots
If B ^ 2-4ac = 0, then the equation has two equal real roots
If B ^ 2-4ac
Let F X be an odd function and f x-f y = f (X-Y) for any XY belonging to R. if x 0 f (1) = - 5, find f (x)
Let F X be an odd function and f x-f y = f (X-Y) for any XY belonging to r if x 0 f (1) = - 2,
It is proved that FX is an odd function
It is proved that FX is a decreasing function on R
3 if f 2x + 5 + F 6-7x ＞ 4, find the value range of X
1) It is proved that if x = 0, - f (y) = f (- y) is an odd function;
2) Proof: let X4
So - 5x + 1113 / 5
1) It is proved that if x = y = 0, we can get f (0) - f (0) = f (0) and f (0) = 0
Let x = 0, we can get f (0) - f (y) = f (- y), that is - f (y) = f (- y), the function is odd;
2) It is proved that let XF (y) be a decreasing function;
3) Start with
1) It is proved that if x = y = 0, we can get f (0) - f (0) = f (0) and f (0) = 0
Let x = 0, we can get f (0) - f (y) = f (- y), that is - f (y) = f (- y), the function is odd;
2) It is proved that let XF (y) be a decreasing function;
3) If we know that the function is an odd function, then f (- 1) = - f (1) = 2, f (- 1) - f (1) = f (- 2) = 4,
∵ f (2x + 5) + F (6-7x) = f (2x + 5) - f (7x-6) = f (- 5x + 11) > 4 = f (- 2), and the function is a decreasing function.
! - 5x + 1113 / 5
Mathematical problems (X & # 178; - 3x + 2) (X & # 178; - x + 2) - 15x & # 178;
(X & \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\35178; + 2 + 2x)
、。 、。 Follow up: it's not a joke, speed
Veda's theorem for quadratic equation of one variable
It is known that the equation x ^ 2 + MX + 4 = 0 and x ^ 2 - (m-2) x-16 = 0 have the same root, and the value of M is the same root
Let the same root of the two equations be m, and the other roots of the former equation and the latter equation be a and B respectively,
∵ a, m are the two roots of the equation x ^ 2 + MX + 4 = 0
∴a+m=-M,am=4
∴(4/m)+m=-M
M=-[(4/m)+m] (1)
∵ B, m are the two roots of the equation x ^ 2 - (m-2) x-16 = 0
∴b+m=M-2,bm=-16
∴(-16/m)+m=M-2 (2)
(1) Substituting (2) to get:
∴(-16/m)+m=-[(4/m)+m]-2
m^2+m-6=0
(m-2)(m+3)=0
∴m=2,-3
m=2,a=4/m=4/2=2,b=-16/m=-16/2=-8
m=-3,a=4/m=-4/3,b=-16/m=16/3
Suppose the same root of the equation is x1, and the different roots are x2 and X3
So (x1) ^ 2 + MX1 + 4 = 0
(X1)^2-(M-2)X1-16=0
If the two equations are subtracted, then X1 = - 10 / (m-1)
So X1 * x2 = 4, X2 + X1 = - M
M = - 4 or M = 13 / 3 can be obtained simultaneously
Then, when m = - 4, X1 = x2 = 2
x1=2,x2=-8
Given that the function FX satisfies FX = - f (- x), and for any x, y belonging to R, there is always FX + FY = f (x + y). If x > 0, FX
1) Let x = a, y = 1, a ∈ R
f(a)+f(1)=f(a+1)
f(a+1)-f(a)=f(1)=-2/3
Is 3x & # 178; - 3 / X-1 = 0 a quadratic equation with one variable
The denominator has unknowns
So it's a fractional equation
It's not a quadratic equation of one variable
no
3 / X is a fraction.
To judge whether an equation is a quadratic equation of one variable, we should first see whether it is an integral equation. If it is, we should sort it out. If it can be sorted out in the form of ax ^ 2 + BX + C = 0 (a ≠ 0), then the equation is a quadratic equation of one variable. There should be an equal sign in it, and the denominator does not contain unknowns.
The above formula can be sorted into 3x & # 178; = 3 / X-1, the third power of X - the square of X-1 = 0 (x is not equal to 1)
This is a cubic equation of one variable, not a quadratic expansion
To judge whether an equation is a quadratic equation of one variable, we should first see whether it is an integral equation. If it is, we should sort it out. If it can be sorted out in the form of ax ^ 2 + BX + C = 0 (a ≠ 0), then the equation is a quadratic equation of one variable. There should be an equal sign in it, and the denominator does not contain unknowns.
The above formula can be sorted into 3x & # 178; = 3 / X-1, the third power of X - the square of X-1 = 0 (x is not equal to 1)
This is a cubic equation of one variable, not a quadratic equation