It is known that P: a = {x | 2A ≤ x ≤ A & # 178; + 1}, Q: B = {x | X & # 178; - 3 (a + 1) x + 2 (3a + 1) ≤ 0}. If P is a sufficient condition for Q, the range of a is obtained

It is known that P: a = {x | 2A ≤ x ≤ A & # 178; + 1}, Q: B = {x | X & # 178; - 3 (a + 1) x + 2 (3a + 1) ≤ 0}. If P is a sufficient condition for Q, the range of a is obtained

If P is a sufficient condition for Q, then p is contained in Q and is transformed into a set problem
The discussion is divided into three situations
(1) When a > 1 / 3, B = {} gives the system of inequalities according to the series of P and Q (the same as below)
(2) a

Two sets a and B are known, set a = {x | X & # 178; - X-2 ≤ 0} and set B = {x | 2A

Set a: X & # 178; - X-2 ≤ 0 (X-2) (x + 1) ≤ 0 - 1 ≤ x ≤ 2
There are two cases: a ∩ B = empty set
(1) On the number axis, B is on the right, 2A > 2, a > 1
(2) On the number axis, B is on the left, a + 3

(X & # 178; + X + 1) (X & # 178; - x + 1) calculate (3a + b-2) (3a-b + 2)

(x²+x+1)(x²-x+1)
=(x²+1)²-x²
=x^4+x²+1
（3a+b-2)(3a-b+2)
=9a²-(b-2)²
=9a²-b²+4b-4

Factoring, adding items, changing elements,
Question 1: X & sup3; - 3x + 2
Question 2: X & sup3; - 48x + 7
Question 3: X & sup3; - 5x & sup2; + 5x-4
Question 4: X (x + 1) (x + 2) (x + 3) + 1
Question 5: (XY-1) & sup2; + (XY-2) (x + y-2xy)

1,x³-3x+2=(x-1)(x²+x-2)2,x³-48x+7=x³-49x+x+7=x(x-7)(x+7)+x+7=(x²-7x+1)(x+7)3,x³-5x²+5x-4=(x³-4x²)-x²+5x-4 =x²(x-4)-(x-4)(x-1) =(x-4)(x²-x+1)...

What is the difference between the method of factorization, the method of collocation and the method of removing and adding terms

The method of adding items is as follows
This method means that one term of a polynomial is separated or filled in two terms (or several terms) which are opposite to each other, so that the original formula is suitable for common factor method, formula method or group decomposition method
bc(b+c)+ca(c-a)-ab(a+b)
=bc(c-a+a+b)+ca(c-a)-ab(a+b)
=bc(c-a)+bc(a+b)+ca(c-a)-ab(a+b)
=bc(c-a)+ca(c-a)+bc(a+b)-ab(a+b)
=(bc+ca)(c-a)+(bc-ab)(a+b)
=c(c-a)(b+a)+b(a+b)(c-a)
=(c+b)(c-a)(a+b)．
Preparation method:
For some polynomials that can not use the formula method, we can match them into a complete square formula, and then use the square difference formula to factorize them. This method is called the matching method. It belongs to a special case of the method of splitting and complementing terms. It should also be noted that the deformation must be carried out under the principle of being equal to the original multinomial formula
For example: X & sup2; + 3x-40 = x & sup2; + 3x + 2.25-42.25 = (x + 1.5) & sup2; - (6.5) & sup2; = (x + 8) (X-5)

In factorization: what are the methods of splitting terms and adding and subtracting terms?

Factorization is the inverse operation of polynomial multiplication. In the operation of polynomial multiplication, several similar terms are often combined into one term or two similar terms with opposite signs are cancelled out to zero. When factoring some polynomials, it is necessary to recover those terms which are combined or cancelled out, that is to say, one of the polynomials is used to decompose some polynomials

（x^2-2x)(x^2-2x+4)+3
(x+1)(x+2)(x+3)(x+4)-120

（x^2-2x)(x^2-2x+4)+3=(x²-2x)²+4(x²-2x)+3;=(x²-2x+1)(x²-2x+3)=(x-1)²(x²-2x+3)(x+1)(x+2)(x+3)(x+4)-120=(x²+5x+4)(x²+5x+6)-120=(x²+5x)²+10(x²+...

1. Factorization of X & sup2; Y & sup2; - X & sup2; - Y & sup2; + 4xy + 1
2. It's a fractional equation
(X-2 out of 3) = 4 - (X & sup2; - 6x + 2 out of 9)
Write down the process, will use word score, try to type out, I will add

1、
The original formula = (X & sup2; Y & sup2; + 2XY + 1) - (X & sup2; - 2XY + Y & sup2;)
=(xy+1)²-(x-y)²
=(xy+x-y+1)(xy-x+y+1)
2、
(x-2)/(x-3)=4-2/(x-3)²
Multiply (x-3) & sup2 on both sides;
(x-2)(x-3)=4(x-3)²-2
x²-5x+6=4x²-24x+36-2
3x²-19x+28=0
(3x-7)(x-4)=0
x=7/3,x=4
The fractional equation needs to be tested
By testing, x = 7 / 3 and x = 4 are the solutions of the equation

-2x^3y+2x^2y^2-1/2xy^3
a^2-8ab+16b^2-c^2-2c-1

1.-2x^3y+2x^2y^2-1/2xy^3
=-2xy(x^2-xy+1/4y^2)
2.a^2-8ab+16b^2-c^2-2c-1
=(a-4b)^2-(c+1)^2

Two factoring problems
(a ^ 2 + 5B ^ 2) / 4 = a ^ 2B + B-1, find the cube root of AB ^ 2
A and B are real numbers
Factorization
2(x-y)-(y-x)^2-(y-x)^3
It's a process. Thank you