The known set a = {x | - 4

The known set a = {x | - 4

B
(x-3a)(x-a)=0
x=3a,x=a
A∩B=∅
So 3A = 6
a=6
So a = 6
A∪B=A
-4

We know the complete set u = R, set a = {x ‖ x ˇ 2-x-60}, C = {x ‖ x ˇ 2-4ax + 3a ˇ 2

A={ -2< X

The complete set u = R, a = {x | x ^ 2-x-6 > 0}, B = {x | x ^ 2 + 2x-8 > 0}, C = {x | x ^ 2-4ax + 3A ^ 2}

It is known from a that x is greater than 3 or X is less than - 2
It is known from B that x is greater than 2 or X is less than - 4
A and B know that x is greater than 3 or X is less than - 4
It is known from C that a is less than X and less than 3A (a is greater than 0), then a is greater than or equal to 3;
Or 3A less than x less than a (a less than 0), then a is less than or equal to - 4
So a should be greater than or equal to 3 or less than or equal to - 4

On ～ of Fractional Inequality
If the numerator and denominator are multiplied by - 1 at the same time, does the inequality need to change sign?

It doesn't need to be changed. When the numerator and denominator are multiplied by any non-zero number, the size of the fraction remains unchanged

If x = 2 is the solution of the inequality X & sup2;; + (a + 1) x + a ≤ 0, find the value range of A

（x+a）（x+1）≤0
If - A1, then the solution set of the inequality is [- A, - 1], and x = 2 is not the solution of the inequality
If - a = - 1, that is, a = 1, then the solution set of the inequality is {- 1}, and x = 2 is not the solution of the inequality
If - a > - 1 is a

Solving the problem of mathematical inequality in grade one of senior high school
It is known that a, B and C are all positive real numbers and satisfy a ^ 2 + B ^ 2 + C ^ 2 = 1
Finding the minimum value of a ^ - 2 + B ^ - 2 + C ^ - 2
The answer is 9

a²+b²+c²=0
So 1 / A & sup2; + 1 / B & sup2; + 1 / C & sup2;
=(1/a²+1/b²+1/c²)(a²+b²+c²)
=1+1+1+b²/a²+c²/a²+a²/b²+c/b²+a²/c²+b²/c²
The arithmetic mean is greater than or equal to the geometric mean
B & sup2 / A & sup2; + C & sup2 / A & sup2; + A & sup2 / B & sup2; + C / B & sup2; + A & sup2 / C & sup2; + B & sup2 / C & sup2; > = 6 (B & sup2 / A & sup2; * C & sup2 / A & sup2; * A & sup2 / B & sup2; * C / B & sup2; * A & sup2 / C & sup2; * B & sup2; * C & sup2;) = 1
So 1 / A & sup2; + 1 / B & sup2; + 1 / C & sup2; > = 1 + 1 + 1 + 6 = 9
So the minimum value is 9

The known function f (x) = 4x / x + A
(1) If a = 1, find the inverse of the function (I've got it, it's - X / x-4)
(2) The inequality f (x) ≥ 1 with respect to X
The key is the second question. Please give me the process. Thank you very much!

4x/(x+a)>=1
4x/(x+a)-1>=0
(3x-a)/(x+a)>=0
(3x-a)(x+a)>=0
（x-a/3)(x+a)>=0
Classification discussion, if
A > 0, then x > A / 3 or x > 1

Factorization formula and examples

There are twelve methods of factorization
The transformation of a polynomial into the product of several integers is called factorization of the polynomial
1. The law of public cause
If each term of a polynomial contains a common factor, then the common factor can be put forward and the polynomial can be transformed into the product of two factors
Decomposition factor X - 2x - x
x -2x -x=x(x -2x-1)
2. Application formula method
Because of the inverse relationship between factorization factor and integral multiplication, if the multiplication formula is reversed, it can be used to factorize some polynomials
Example 2: decomposition factor A + 4AB + 4B
a +4ab+4b =（a+2b）
3. Group decomposition method
To decompose the polynomial am + an + BM + BN, we can first divide its first two terms into a group, and propose the common factor A, divide its last two terms into a group, and propose the common factor B, so as to obtain a (M + n) + B (M + n), and propose the common factor M + N, so as to obtain (a + b) (M + n)
Example 3: decomposition factor M + 5n-mn-5m
m +5n-mn-5m= m -5m -mn+5n
= (m -5m )+(-mn+5n)
=m(m-5)-n(m-5)
=(m-5)(m-n)
4. Cross multiplication
If a × B = m, C × d = q and AC + BD = P, then the polynomial can be factorized as (AX + D) (BX + C)
Example 4. Factorization 7x - 19x - 6
Analysis: 1 - 3
7 2
2-21=-19
7x -19x-6=（7x+2）(x-3)
5. Matching method
For those polynomials that can't use formula method, some can be factorized by matching them into a complete square formula and then using the square difference formula
Example 5: Factorization x + 3x-40
Solution x + 3x-40 = x + 3x + () - () - 40
=(x+ ) -( )
=(x+ + )(x+ - )
=(x+8)(x-5)
6. Method of removing and adding items
The polynomial can be divided into several parts, and then factorized
Example 6: decomposition factor BC (B + C) + Ca (C-A) - AB (a + b)
bc(b+c)+ca(c-a)-ab(a+b)=bc(c-a+a+b)+ca(c-a)-ab(a+b)
=bc(c-a)+ca(c-a)+bc(a+b)-ab(a+b) =c(c-a)(b+a)+b(a+b)(c-a) =(c+b)(c-a)(a+b)
7. Substitution method
Sometimes when factoring, you can choose the same part of the polynomial to be replaced by another unknown, then factoring, and finally converting back
Example 7: Factorization factor 2x - X - 6x - x + 2
2x -x -6x -x+2=2(x +1)-x(x +1)-6x
=x [2(x + )-(x+ )-6
Let y = x +, x [2 (x +) - (x +) - 6]
= x [2(y -2)-y-6]
= x (2y -y-10)
=x (y+2)(2y-5)
=x (x+ +2)(2x+ -5)
= (x +2x+1) (2x -5x+2)
=(x+1) (2x-1)(x-2)
8. Root seeking method
Let the polynomial f (x) = 0, find its roots as X, x, x Then the polynomial can be factorized as f (x) = (x-x) (x-x) (x-x) (x-x )
Example 8: Factorization factor 2x + 7x - 2x - 13X + 6
Let f (x) = 2x + 7x - 2x - 13X + 6 = 0
Through the comprehensive division, we can know that f (x) = 0, the root is, - 3, - 2,1
Then 2x + 7x - 2x - 13X + 6 = (2x-1) (x + 3) (x + 2) (x-1)
9. Image method
Let y = f (x), make the image of function y = f (x), find the intersection x, x, x Then the polynomial can be factorized as f (x) = f (x) = (x-x) (x-x) (x-x) (x-x )
Example 9: Factorization x + 2x - 5x-6
Let y = x + 2x - 5x-6
See the figure on the right. The point of intersection with x-axis is - 3, - 1,2
Then x + 2x - 5x-6 = (x + 1) (x + 3) (X-2)
10. Principal component method
First, select a letter as the main element, then arrange the items according to the letter number from high to low, and then factorize them
Example 10: decomposition factor A (B-C) + B (C-A) + C (a-b)
Analysis: this problem can choose a as the main element, and arrange it from high to low
a (b-c)+b (c-a)+c (a-b)=a (b-c)-a(b -c )+(b c-c b)
=(b-c) [a -a(b+c)+bc]
=(b-c)(a-b)(a-c)
11. Using the special value method
Substituting 2 or 10 into x, finding the number P, decomposing the number P into prime factors, properly combining the prime factors, writing each factor after combination into the form of sum and difference of 2 or 10, and reducing 2 or 10 to x, the factorization formula is obtained
Example 11: Factorization x + 9x + 23x + 15
Let x = 2, then x + 9x + 23x + 15 = 8 + 36 + 46 + 15 = 105
105 is decomposed into the product of three prime factors, that is, 105 = 3 × 5 × 7
Note that the coefficient of the highest term in the polynomial is 1, while 3, 5, and 7 are x + 1, x + 3, and X + 5, respectively, when x = 2
Then x + 9x + 23x + 15 = (x + 1) (x + 3) (x + 5)
12. Undetermined coefficient method
Firstly, the form of factorization factor is judged, then the letter coefficient of the corresponding integral is set, and the letter coefficient is calculated, so as to decompose the polynomial factor
Example 12. Factorization factor X - X - 5x - 6x - 4
Analysis: it is easy to know that this polynomial has no first-order factor, so it can only be decomposed into two quadratic factors
Let X - X - 5x - 6x - 4 = (x + ax + b) (x + CX + D)
= x +(a+c)x +(ac+b+d)x +(ad+bc)x+bd
So it's a good solution
Then x - X - 5x - 6x-4 = (x + X + 1) (x - 2X-4)

Factorization method, seeking explanation

In factorization, sometimes in order to use formula, one term in the formula is divided into two terms, so as to use formula method to decompose. This method is called term splitting method

Factorization method
x^3+2x-3
x^3-8

x^3+2x-3=x^3-x^2+x^2+2x-3=x^2(x-1)+(x-1)(x+3)=(x-1)(x^2+x+3)
x^3-8 =(x-2)(x^2+2x+4)