Let u = R, a = {x | 3x-4 | > 5}, B = {x | x ^ 2 + 4x0} (1) B and C (2) (C intersection a) intersection B

Let u = R, a = {x | 3x-4 | > 5}, B = {x | x ^ 2 + 4x0} (1) B and C (2) (C intersection a) intersection B

A={x|x3}B={x|-7
B∪C={x|-7
The derivation and conclusion of Weida theorem
What are the conditions~
Let ax ^ 2 + BX + C = 0 (a is not 0)
When △≥ 0
X1 = (- B + root △) / 2A x2 = (- B - root △) / 2A
So X1 + x2 = [(- B + radical △) / 2A] + [(- B - radical △) / 2A]
=-2b/2a
=-b/a
Similarly, x1x2 = [(- B + root △) / 2A] * [(- B - root △) / 2A]
=[(-b)^2-(b^2-4ac)]/4a^2
=4ac/4a^2
=c/a
It is known that the function f (x) is an odd function defined on R?
On the negative half axis of X, move the image of 2 ^ x up one unit. On the positive half axis of X, just make the image of the negative half axis symmetrical about the origin. F (0) = 0
Find the solution set of the following inequality: (1) 4x ^ 2-4x ＞ 15 (2) 13-4x ^ 2 ＞ 0 (3) x ^ 2-3x-1
Find the solution set of the following inequality:
(1)4x^2－4x＞15 (2)13－4x^2＞0
(3)x^2－3x－10＞0 (4)x(9－x)＞0
(1) 4X ^ 2-4x ＞ 15 (2x + 3) (2x-5) ＞ 0 x ＜ - 3 / 2 or X ＞ 5 / 2
(2) 13-4x ^ 2 ＞ 0, X is greater than the root of minus two, 13 is less than the root of two, 13
(3) X ^ 2-3x-10 ＞ 0 (x + 2) (X-5) ＞ 0 x ＜ - 2 or X ＞ 5
(4)x(9－x)＞0 ﹙x＋3﹚﹙x－3﹚＜0 －3＜x＜3
What is x ^ 2? Square of X? That's a good solution. I'm not good at math. I just passed the college entrance examination
How to deduce the Veda theorem of cubic function
As we all know, for the univariate quadratic equation AX ^ 2 + BX + C = 0, (a ≠ 0) two x1, X2 have the following relation: X1 + x2 = - B / a x1x2 = C / a|x1-x2 | = √ △ / |a | for the third one, the proof is very simple, that is, the difference root sign depends on the square of Formula 1 and the multiplication of formula 2 by 4
It's similar:
ax^3+bx^2+cx+d=a(x-x1)(x-x2)(x-x3)=a[x^3-(x1+x2+x3)x^2+(x1x2+x2x3+x1x3)x-x1x2x3]
The contrast coefficient is as follows:
x1+x2+x3=-b/a
x1x2+x2x3+x1x3=c/a
x1x2x3=-d/a
It is known that the function f (x) is an odd function defined on R. when x ≥ 0, f (x) = x (1 + x), the image of function f (x) is drawn and the analytic expression of function f (x) is obtained
∵ when x ≥ 0, f (x) = x (1 + x) = (x + 12) 2-14, f (x) is an odd function defined on R, ∵ when x < 0, - x > 0, f (- x) = - x (1-x) = (X-12) 2-14 = - f (x), ∵ f (x) = - (X-12) 2 + 14 ∵ f (x) = (x + 12) & nbsp; 2-14x ≥ 0 - (X-12) & nbsp; 2 + 14x < 0
Given a = {x | x ^ 2-3x + 2 = 0, X ∈ r}, B = {x | 0
Solve the equation x ^ 2-3x + 2 = 0 to get x = 1 or x = 2
Then a = (1,2)
If 0 ＜ x ＜ 5, X ∈ n, x = 1,2,3,4
Then B = (1,2,3,4)
B includes A. is C required to be included by B and include a?
So there are (1,2,3) (1,2,4) two
How to deduce the Veda theorem of cubic equation
ax^3+bx^2+cx+d
=How is a (x-x1) (x-x2) (x-x3) derived?
This step is not deductive
On the left is the general form of cubic equation
On the right is the zero form of the cubic equation
Because x1, X2, X3 are the roots of the equation, then we can continue to deduce
It is known that the function f (x) is an odd function defined on R. when x ≥ 0, f (x) = x (1 + x), the image of the function f (x) is drawn and the analytic expression of the function is obtained
On the definition of odd function R (f)
Then f (x) = - f (- x)
Let x0 be brought in
f(-x)=-x(1-x)
Then f (x) - f (- x) = x (1-x)
Then f (x) = x (1 + x) x > = 0
f（x）＝x（1－x） x
When x ≥ 0, f (x) = x (1 + x),
X0,
f(-x)=-x(1-x)
f(x)=-f(-x)=x(1-x)
Given the set a = {x | x2-3x + 5 = 0}, B = {x | (x + 1) 2 (x2 + 3x-4) = 0}, and a ⊊ P ⊆ B, find the set P satisfying the condition
From (x + 1) 2 (x2 + 3x-4) = 0, the solution of x = - 1, or 1, or - 4, the set B = {- 4, - 1, 1}. And ⊊ a ⊊ P ⊆ B, the set P is a nonempty subset of the set B, P = {- 4}, {- 1}, {- 4, - 1}, {- 4, 1}, {- 1, 1}, {