Given the set a = {x | - 1 ≤ x ≤ 3}, B = {y | - y = x & # 178;, X ∈ a}, C = {y | - y = 2x + A, X ∈ a}, if C & # 8838; B, Finding the value range of real number a

Given the set a = {x | - 1 ≤ x ≤ 3}, B = {y | - y = x & # 178;, X ∈ a}, C = {y | - y = 2x + A, X ∈ a}, if C & # 8838; B, Finding the value range of real number a

Because x ∈ a, so - 2 ≤ 2x ≤ 6
So - 2 + a ≤ 2x + a ≤ 6 + A
That is - 2 + a ≤ y ≤ 6 + a
B ：0≤y≤9
Because C is a subset of B
So - 2 + A is greater than or equal to 0 and 6 + A is less than or equal to 9
SO 2 ≤ a ≤ 3

Given the set a = {x | y = x & # 178; - 2x-3, X ∈ r}, B = {y | y = x & # 178; - 2x-3, X ∈ r},
Set C = {(x, y) | y = x & # 178; - 2x-3, x, y ∈ r}, d = {(x, y) | y + 3 / X-2 = 1, x, y ∈ r}, find a ∩ B, B ∩ C, C ∩ D

From the known: a ∈ R, B = {y ≥ - 4},
So a ∩ B = {x ≥ - 4};
Because the B set line, the C set is the point
So B ∩ C is an empty set;
The simultaneous C and D equations are solved

Given the set a = {y | y = x2 + 2x, - 2 ≤ x ≤ 2}, B = {x | x2 + 2x-3 ≤ 0}, taking any element a in the set a, then the probability of a ∈ B is___ ．

A = {y | y = x2 + 2x, - 2 ≤ x ≤ 2} = {y | - 1 ≤ y ≤ 8}, B = {x | - x = x2 + 2x-3 ≤ 0} = {x | - 3 ≤ x ≤ 1}, if a ∈ B, then - 1 ≤ a ≤ 1} according to the probability formula of geometric probability, if any element a is taken from the set a, then the probability p of a ∈ B is 1 - (- 1) 8 - (- 1) = 29, so the answer is: 29

Factorization-81 + A ^ 4
-81+a^4
4（x+1）^2-9

Original = (A & # 178; + 9) (A & # 178; - 9)
=(a²+9)(a+3)(a-3)
The original formula = [2 (x + 1)] ² - 9
=(2x+2)²-3²
=(2x+2+3)(2x+2-3)
=(2x+5)(2x-1)

Solving factorization problems
It is known that there are factors (x-1) and (X-2) to find m, n

For example: A & sup2; + B & sup2; - C & sup2; - 4D & sup2; - 2Ab + 4CD X & sup2; + 2XY + Y & sup2; - 2x-2y + 1 = 0

a²+b²-c²-4d²-2ab+4cd
=a²+b²-2ab-（c²+4d²-4cd)
=(a-b)²-（c-2d）²
=（a-b+c-2d）（a-b-c+2d）
X & sup2; + 2XY + Y & sup2; - 2x-2y + 1 = 0 (first, the right side of the factorization problem will not equal 0) this is the equation
x²+2xy+y²-2x-2y+1=(x+y)²-2（x+y）+1=（x+y-1）²
Go back and have a look, square difference formula and complete square formula!

Solve some factorization problems
1.2x²-4x 2.8m²n+2mn 3.a²x²y-axy² 4.3x³-3x²-9x 5.-24x²-12xy²+28y³ 6.-4a³b³+6a²b-2ab 7.-2x²-12xy²+8xy³ 8.-3ma³+6ma²-12ma

1.2x²-4x
Extracting common factors
=2x(x-2)
2.8m²n+2mn
Extracting common factors
=2mn(4m+1)
3.a²x²y-axy²
Extracting common factors
=axy(ax-y)
4.3x³-3x²-9x
Extracting common factors
=3x(x²-x-3)
5.-24x²-12xy²+28y³
Extracting common factors
=-4(6x²+3xy²-7y³)
6.-4a³b³+6a²b-2ab
Extracting common factors
=-2ab(2a²b²-3a+1)
7.-2x²-12xy²+8xy³
Or extract the common factor
=-2x(x+6y²-4y³)
8.-3ma³+6ma²-12ma
The common factor is also extracted
=-3ma(a²-2a+4)

1.（2x+y）^2-（x+2y）^2
2.（p-4）（p+1）+3p
3.3ax^2-3ay2
4.12a（x^2+y^2）-18b（x^2+y^2）
5.-a^5+a
6.3（m+n）^2-27n^2
7.25（x+y）^2-9（x-y）^2
There is another factorization, and then calculate π R1 ^ 2 + π R2 ^ 2 + π R3 ^ 2, where R1 = 4, R2 = 8, R3 = 10, π is 3.14

1.（2x+y）^2-（x+2y）^2
=(2x+y+x+2y)(2x+y-x-2y)
=(3x+3y)(x-y)
=3(x+y)(x-y)
2.（p-4）（p+1）+3p
=p²-3p-4+3p
=p²-4
=(p+2)(p-2)
3.3ax^2-3ay2
=3a(x²-y²)
=3a(x+y)(x-y)
4.12a（x^2+y^2）-18b（x^2+y^2）
=6(x²+y²)(2a-3b)
5.-a^5+a
=-a(a^4-1)
=-a(a²+1)(a²-1）
=-a(a²+1)(a+1)(a-1)
6.3（m+n）^2-27n^2
=3[(m+n)²-(3n)²]
=3(m+n+3n)(m+n-3n)
=3(m+4n)(m-2n)
7.25（x+y）^2-9（x-y）^2
=[5(x+y)]²-[3(x-y)]²
=(5x+5y+3x-3y)(5x+5y-3x+3y)
=(8x+2y)(2x+8y)
=8(4x+y)(x+4y)
Supplementary questions:
There is another factorization, and then calculate π R1 ^ 2 + π R2 ^ 2 + π R3 ^ 2, where R1 = 4, R2 = 8, R3 = 10, π is 3.14
πR1^2+πR2^2+πR3^2
=π(R1^2+R2^2+R3^2)
=3.14(16+64+100)
=565.2

Factorization calculation problem
Factorization:
（1） 2ax-10ay+5by+6x
（2） 1-a^2-ab-1/4b^2
（3） x^5y-9xy^5
（4） 4a-5a^5
（5） 2x^2-4x+1
（6） 4y^2+4y-5
(7) 3x^2-7x+2
I want to talk about the process. What formula is used? Thank you!

2.1-（a^2+ab+1/4b^2)=1-(a+1/2b)^2=(1+a+1/2b)(1-a-1/2b)
3.xy(x^4-9y^4)=xy(x^2+3y^2)(x^2-3y^2)
4.a(4-5a^4)
7(3x-1)(x+2)
Is your first question wrong? There is no integer solution in question 5.6, is there a problem? There is no good integer solution in question 4, so you can look at it again

How to solve this problem
X ^ 2 + y ^ 2 + 2x-4y + 5 = 0, find X-Y

x^2+y^2+2x-4y+5=x^2+2x+1+y^2-4y+4=(x+1)^2+(y-2)^2=0
x=-1
y=2
x-y=-3