Given the complete set u = {1,2,3,4,5}, if a ∪ B = u, a ∩ B ≠ 0, a ∩ (∁ UB) = {1,2}, try to write a set of a and B satisfying the condition

Given the complete set u = {1,2,3,4,5}, if a ∪ B = u, a ∩ B ≠ 0, a ∩ (∁ UB) = {1,2}, try to write a set of a and B satisfying the condition

A ∪ B = u and a ∩ (cub) = {1,2}, so {1,2} 8838; a, 3 8838; a, 3 ⊆ a, 3 ⊆ B, 4 ⊎ B = u and a ∩ (cub) = {1,2}, so {1,2} 8834\8838; a, so {1,2} a 88383888383838,1,2,2,2,2,3,3,3,3,3,3,3,4,4,4,4,4,4,4,4,4,4,4,4,4,4,5,5}, or {1, 2, 3, 4} or {1, 2

The function f (x) defined on R satisfies f (M + N2) = f (m) + 2 [f (n)] 2, where m, n ∈ R and f (1) ≠ 0=______ ．

According to the title, f (2013) = f (2012 + 12) = f (2012) + 2 [f (1)] 2, f (2012) = f (2011) + 2 [f (1)] 2, f (2011) = f (2010) + 2 [f (1)] 2, f (2010) = f (2009) + 2 [f (1)] 2 F (2) = f (1) + 2 [f (1)] 2, so f (2013) = f (1) + 2 [f (1)] 2 × 2012 = 4024 [f (1)] 2 + F (1), so the answer is 4024 [f (1)] 2 + F (1)

Let f (x) be the set of functions defined on R, M = {x | f (x) = x}, n = {x | f (f (x)) = x}
Prove that M is contained in n
If f (x) is an increasing function on R, judge whether M = n holds and prove your conclusion

27. Let f (x) be a set of functions defined on R, M = {x | f (x) = x}, n = {x | f (f (x)) = x} 1. Prove that M is contained in N2. If f (x) is an increasing function on R, judge whether M = n holds, and prove your conclusion: 1. Let x ∈ m, then f (x) = x, ｜ f [f (x)] = f (x) = x, ｜ x ∈ n, ｜ m is contained in N. 2. Let y ∈ n, X = f (y)}

Several factoring problem will not do, please help! To detailed process!
(1) The third power of 32A - the square of 50ab
(2) The square of 9 / M - the fourth power of 256 / n
(3) The 2nth power of 4-x
(4) The square of (a + b) - 1
(5) The fourth power of 1-16a and the fourth power of B

（a-b）^3-2ab（a-b）
x（x+y）（x-y）-x（x+y）^2
a^2（x-2a）^2-a（2a-x）^2
（b+c）x+（c+a）x+（a+b）x
a^3（b+c-d）+a^2b（c+d-a）-a^2c（d+a+b）

（a-b）^3-2ab（a-b）
=(a-b)[(a-b)²-2ab]
=（a-b)(a²-4ab+b²)
x（x+y）（x-y）-x（x+y）^2
=(x+y)(x²-xy-x²-xy)
=-2xy(x+y)
a^2（x-2a）^2-a（2a-x）^2
=a(x-2a)²(a-1)
（b+c）x+（c+a）x+（a+b）x
=x(2a+2b+2c)
=2x(a+b+c)
a^3（b+c-d）+a^2b（c+d-a）-a^2c（d+a+b）
=a²(ab+ac-ad+bc+bd-ab-dc-ac-bc)
=a²(-ad+bd)
=a²d(b-a)

1、x³+3x²-4
2、x³+9x²+26x+24
3、（2x²-3x+1）²-22x+33x-1
4、（x+y）³+2xy(1-x-y)-1
5、(x+3)(x²-1)(x+5)-20

1. In this paper, we are going to take the item method: x-179; (x-179; x-x-178; (x-x-178, + 4x-178, + 4x-178; (x-1) + 4 (x + 1) (x-1) + 4 (x + 1) (x-1) (x-1) (x-1) (x-179; (x-178; (x-1) + 4 (x-1) (x-1) (x-178; x-1) (x-178; x-1) (x-178; (x-1) (x-178; (x-1) (x-178; (x-1) (x-178; x-1) (x-178; (x-1) (x-178; (x-178; (x-1) (x-1) (x-178; (x-178; (x-178; (x-178; (x-1) (x-178; (x-1) (x-178; x-178; x-178; x-178; x-178; x-178; x-178; x-178; x-178; + 12 (x + 2) = (x + 2) (X & # 178; + 7x + 12) = (x + 2) (x + 3) (x + 4). 3. (suppose element method): let 2x & # 178; - 3x + 1 = y, then the original formula = y & # 178; - 11y + 10 = (Y-10) (Y-1), ｜ original formula = (2x & # 178; - 3x + 1-10) (2x & # 178; - 3x + 1-1) = x (2x + 3) (x-3) (2x-3). 4. (cubic formula): original formula = (x + y) &# 179; In this paper, we will (x + y-1-2xy (x + X + Y-1) (x + y-1-2xy (x + Y-1) and (x + X + y + 1) 2XY (x + X + Y-1) (x + Y-1) (x + Y-1) (x + Y-1) (x + Y-1) (x + Y-1) (x-1) (x-1) (x-1) (x-1) (x-1) (x + 5) (x + 5) (x + 5) (x + 5) (x + 5) (x + 5) (x + 5) (x + X + 5) (x + X-5) (x (x-x-5) (X-5) (X-5) (x-x-178 (X-5) (x) (X-5) (X-5) (x-178 (x-x-5) (x-178) (X-5) (x-178) (X-5) (x-178) (X-5) (X-5) (x-178) (x-178) (x-178) (x-178, x- 20 = y & # 178; －36=﹙y＋6﹚﹙y－6﹚=﹙x²＋4x＋5﹚﹙x²＋4x－7﹚

1. Factorization: A ^ 2 (a-b) + (B-A) given x + y = 2, xy = 2 / 3, find: (X-Y) ^ 2

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

1.a^4-15a^2+9
2.x^3-48x+7
3.4x^2-（x^2+1）^2
4.（5x^2-13y^2）^2-16（x^2-3y^2）^2
[Note: decompose thoroughly / write a more detailed process / I will add scores]

1. The original formula = (a ^ 4-6a ^ 2 + 9) - 9A ^ 2
=(a^2-3)^2-9a^2
=(a^2-3a-3)(a^2+3a-3)
2. The original formula = (x ^ 3 + 7x ^ 2) - (7x ^ 2 + 49x) + (x + 7)
=(x+7)(x^2-7x+1)
3. Original formula = (2x + x ^ 2 + 1) (2x-x ^ 2-1)
=-(x+1)^2(x-1)^2
4 original = (5x ^ 2-13y ^ 2 + 4x ^ 2-12y ^ 2) (5x ^ 2-13y ^ 2-4x ^ 2 + 12Y ^ 2)
=(3x+5y)(3x-5y)(x+y)(x-y)

Factorization, I want speed
Factorization: x ^ 2 + X-5
10: The answer came out before 00,
If x ^ 2 + X-5 = 0, then x =?

x^2+x-5=0,
The root is x = (- 1 ± √ 21) / 2,
x^2+x-5=[x-(-1+√21)/2)][x-(-1-√21)/2]

Several factorization problems
The square of 16-x, the square of Y
7x squared - 63
Square of (x + x) square of (x + 4Y)
Square of 25 (M + n) - square of 4 (m-n)
Square of (x + y + Z) - (X-Y-Z)
There is a specific process to strive for

1 16-x²y²=（4-xy）（4+xy）
2 7x²-63=7（x²-9）=7（x+3）（x-3）
3 （x²+x)²-（x+4y）²
=（x²+x-x-4y）（x²+x+x+4y）
=（x²-4y）（x²+2x+4y）
4 25（m+n）²-4（m-n）²
=（5m+5n）²-（2m-2n）²
=（5m+5n-2m+2n）（5m+5n+2m-2n）
=（3m+7n）（7m+3n）
5 （x+y+z）²-（x-y-z）²
=（x+y+z-x+y+z）（x+y+z+x-y-z）
=（2y+2z）（2x）
=4（y+z）x