If for any two elements in a set, the result after some operation is still in the set, then the set is said to be closed to this operation. Given that a = {0,1}, B = {y | y = m + n √ 2, m, n ∈ Z}, try to judge whether a and B are closed to the four operations of addition, subtraction, multiplication and division. Why?

If for any two elements in a set, the result after some operation is still in the set, then the set is said to be closed to this operation. Given that a = {0,1}, B = {y | y = m + n √ 2, m, n ∈ Z}, try to judge whether a and B are closed to the four operations of addition, subtraction, multiplication and division. Why?

A was investigated
Addition: ∵ 1 + 1 = 2 &; a, ∥ unclosed
Subtraction: ∵ 0-1 = - 1 &; a, ∥ not closed
Multiplication: ∵ 0 × 0 = 0 × 1 = 0 ∈ A and 1 × 1 = 1 ∈ a, ∵ closed
Division: ∵ 0 △ 0 &; a, ∵ unclosed
B was investigated
Let P, Q ∈ B and P = a + B √ 2, q = C + D √ 2
Addition: ∵ P + q = (a + C) + (B + D) √ 2 ∈ B, ∵ closed
Subtraction: ∵ P-Q = (A-C) + (B-D) √ 2 ∈ B, ∵ closed
Multiplication: ∵ PQ = (AC + 2bd) + (AD + BC) √ 2 ∈ B, ∵ closed
Division: P / Q = [(ac-2bd) + (BC AD) √ 2] / [C & # 178; - 2D & # 178;]
If C & # 178; - 2D & # 178; = 0, then C = √ 2D, at least one of C and D does not belong to Z or both are 0
The contradiction between the former and the meaning of the title can be eliminated, and the latter can not be closed
A 0-1 = - 1 subtraction does not close other closures
B calculate (a + B √ 2) / (c + D √ 2) and (a + B √ 2) * (c + D √ 2), and the results are in the form of M + n √ 2
It is easy to see that (a + B √ 2) - (c + D √ 2) and (a + B √ 2) + (c + D √ 2) also have the form of M + n √ 2, so B is closed to all four operations
Read the meaning first
Set a has only two elements 0 and 1
Then add 0 + 1 = 1, 1 belongs to a, so it can
Note that 1-0 = 1 is OK, but the emphasis of closed set is arbitrary, because 0-1 = - 1 is not OK
Multiplication is the same as 1x0 = 0
It is meaningless to divide by 1 / 0 without considering 0 / 1 = 0
For set B
This is an abstract collection... Expanded
Read the meaning first
Set a has only two elements 0 and 1
Then add 0 + 1 = 1, 1 belongs to a, so it can
Note that 1-0 = 1 is OK, but the emphasis of closed set is arbitrary, because 0-1 = - 1 is not OK
Multiplication is the same as 1x0 = 0
It is meaningless to divide by 1 / 0 without considering 0 / 1 = 0
For set B
This is an abstract set and can be solved
Let Y1 = M1 + √ 2n1
y2=m2+√2n2
By adding Y1 + y2 = M1 + √ 2n1 + M2 + √ 2n2 = (M1 + m2) + √ 2 (N1 + N2), we can
The subtraction is the same as y1-y2 = (m1-m2) + √ 2 (n1-n2)
Multiply Y1 x y2 = (m1m2 + 2n1n2) + √ 2 (m1n2 + m2n1), so you can
Except Yi / y2 = (M1 + √ 2n1) / (M2 + √ 2n2)
Multiplication of numerator and denominator (M2 - √ 2n2)
We get Yi / y2 = (m1m2 + 2n1n2) / (M2 ^ 2-2n2 ^ 2) + √ 2 (m1n2 + m2n1) / (M2 ^ 2-2n2 ^ 2)
For (m1m2 + 2n1n2) / (M2 ^ 2-2n2 ^ 2) and √ 2 (m1n2 + m2n1) / (M2 ^ 2-2n2 ^ 2)
It can't be reduced to the simplest integer
Therefore, it does not meet the requirements of set B, so it can not be folded
Given the set a = {X / X & sup2; + PX + q = 0}, B = {QX & sup2; + PX + 1 = 0}, at the same time satisfy that a intersection B is not empty set, a ∩ CRB = {2}, where P and Q are real numbers not equal to 0, find P and Q
Let the element in a set be a, and the element in B set be B, then B: QB & sup2; + Pb + 1 = 0 divide both sides of the equation by B & sup2; to get (1 / b) & sup2; + P (1 / b) + q = 0A: a & sup2; + PA + q = 0. It can be seen that the elements in a set and B set are opposite to each other. From the intersection of a and B, we get x = 1 / x, and the solution is x = ± 1, so the elements in a are {2,1} or {2
One hundred and eleven
Given the complete set u = {x-x ^ 2 + 3x-2 ≤ 0}, set a = {x-x-2 > 1}, set B = {x (x + 1) / (X-2) ≥ 0}, find: (1) a ∩ B (2) a ∪ B (3) a ∩ cub (4) CUA ∪ B
U={x|-x²+3x-2≤0}={x|x²-3x+2≥0}={x|(x-1)(x-2)≥0}={x|x≥2 or x≤1}A={x| |x-2|>1}={x|x-2>1 or x-23 or x2 or x≤-1}(1),A∩B={x| x≤-1or x>3 }(2),A∪B={x| x>2 or x
The complete set is greater than or equal to 2 or less than or equal to 1
A is greater than 3 or less than 1
B is less than or equal to negative 1 or greater than 2
Then ask for it yourself.
It's very simple. You have to think for yourself to improve
x> 3 or x = 2 or X
To solve a mathematical problem about factorization
Factorization: x-2x & # 178;
x-2x²=x(1-2x)
Original formula = x (1-2x)
Factorization: x-2x & # 178; = x (1-2x)
X*（1-2X）
Given the square n - () + () of the polynomial 2m, fill in the appropriate polynomial (both contain two terms) in each bracket to make the result 2 after merging the similar terms
Please write a complete polynomial like this
2m^2n-（3m^2-1 ）+（mn^2+1 ）=2
Find 20 factorization problems and the answer process
There is another one, please accept
What about the title???
(A's square + 2Ab + B's Square) - (A's Square - 2Ab + B's Square) first remove the brackets, and then merge the similar terms
（a²+2ab+b²）-（a²-2ab+b²）
Remove brackets = A & # 178; + 2Ab + B & # 178; - A & # 178; + 2ab-b & # 178;
=4ab
（a²+2ab+b²）-（a²-2ab+b²）
Remove brackets = A & # 178; + 2Ab + B & # 178; - A & # 178; + 2ab-b & # 178;
=4ab
Factorization in grade two of junior high school
A ^ 4 + A ^ 2-A · B ^ 3-AB = ("-" is the minus sign and "·" is the multiplication sign. In order to distinguish (AB) ^ 2 from a ^ 2 multiplied by B)
xy—xz—y^2+2yz=
x^2—xy+4x—4y=
x^3+3·x^2—4x—12=
x^3+6x^2+11x+6=
2. X (^ 3n) - 12. X (^ 2n) · y (^ 2) + 18. X (^ n) · y (^ 4) = (it's a bit messy here, so the powers are all enclosed in brackets)
（ax+by）^2+（ay—bx）^2+c^2 · x^2+c^2 · y^2
（x^2+y^2）^3+（z^2—x^2）^3—（y^2+z^2）^3
x^6+64·y^6+12x（^2）y(^2)—1
It's a bit messy. In fact, it's on paper. Otherwise, it's only for tonight! The sooner the better
a^4＋a^2—a·b^3—ab=a(a³-b³)+a(a-b)=a(a-b)(a²+ab+b²)+a(a-b)=a(a-b)(a²+ab+b²+1)
xy—xz—y^2+2yz=x(y-z)-y(y-z)=(y-z)(x-y)
x^2—xy+4x—4y=x(x-y)+4(x-y)=(x-y)(x+4)
x^3+3·x^2—4x—12=x²(x+3)-4(x+3)=(x+3)(x²-4)=(x+3)(x-2)(x+2)
x^3+6x^2+11x+6=x³+6x²+9x+2x+6=(x+3)²+2(x+3)=(x+3)(x+5)
2·x(^3n) — 12·X(^2n ）· y（^2）+18·x(^n) · y(^4)=
=2x^n(x^2n-6x^ny^2+9y^4
=2x^n(x^n-3y²)²
（ax+by）^2+（ay—bx）^2+c^2 · x^2+c^2 · y^2
=a²x²+2abxy+b²y²+a²y²-2abxy+b²x²+c²x²+c²y²
=a²x²+b²y²+a²y²+b²x²+c²x²+c²y²
=a²(x²+y²)+b²(x²+y²)+c²(x²+y²)
=(x²+y²)(a²+b²+c²)
（x^2+y^2）^3+（z^2—x^2）^3—（y^2+z^2）^3
=(x²+y²+z²-x²)[(x²+y²)²-(x²+y²)(z²-x²)+(z²-x²)²]-(y²+z²)³
=(y²+z²)(x^4+2x²y²+y^4-x²z²-y²z²+x^4+x²y²+z^4-2x²z²+x^4-y^4-2y²z²-z^4)
=(y²+z²)(3x^4+3x²y²-3x²z²-3y²z²)
=3(y²+z²)(x^4+x²y²-x²z²-y²z²)
=3(y²+z²)[x²(x²+y²)-z²(x²+y²)]
=3(y²+z²)(x²+y²)(x²-z²)
x^6+64·y^6+12x（^2）y(^2)—1
a^4+a^2-ab^3-ab
=a(a³-b³)+a(a-b)
=a(a-b)(a²+b²+ab+1)
xy—xz—y^2+yz
=xy-xz-y²+yz
=x(y-z)-y(y-z)
=(y-z)(x-y);
x^2—xy+4x—4y
=x(x+4)-y(x+4)
=(x -... Expansion)
a^4+a^2-ab^3-ab
=a(a³-b³)+a(a-b)
=a(a-b)(a²+b²+ab+1)
xy—xz—y^2+yz
=xy-xz-y²+yz
=x(y-z)-y(y-z)
=(y-z)(x-y);
x^2—xy+4x—4y
=x(x+4)-y(x+4)
=(x-y)(x+4);
x^3+3·x^2—4x—12
=x²(x+3)-4(x+3)
=(x²-4)(x+3)
=(x-2)(x+2)(x+3)；
x^3+6x^2+11x+6
=x³+6x²+8x+3x+6
=x(x²+6x+8)+3(x+2)
=x(x+2)(x+4)+3(x+2)
=(x+2)(x²+4x+3)
=(x+2)(x+1)(x+3)
2·x(^3n) — 12·X(^2n ）· y（^2）+18·x(^n) · y(^4)
=2x^(3n)-6x^(2n)y²-6x^(2n)y²+18x^ny^4
=2x^(2n)(x^n-3y²)-6x^ny²(x^n-3y²)
=(2x^(2n)-6x^ny²)(x^n-3y²)
=2x^n(x^n-3y²)²;
（ax+by）^2+（ay—bx）^2+c^2 · x^2+c^2 · y^2
=a²x²+b²y²+a²y²+b²x²+c²x²+c²y²
=(a²+b²+c²)x²+(a²+b²+c²)y²
=(a²+b²+c²)(x²+y²)
（x^2+y^2）^3+（z^2—x^2）^3—（y^2+z^2）^3
=(x²+y²+z²-x²)[(x²+y²)²-(x²+y²)(z²-x²)+(z²-x²)²]-(y²+z²)³
=(y²+z²)(x^4+2x²y²+y^4-x²z²-y²z²+x^4+x²y²+z^4-2x²z²+x^4-y^4-2y²z²-z^4)
=(y²+z²)(3x^4+3x²y²-3x²z²-3y²z²)
=3(y²+z²)(x^4+x²y²-x²z²-y²z²)
=3(y²+z²)[x²(x²+y²)-z²(x²+y²)]
=3(y²+z²)(x²+y²)(x²-z²)
There's something wrong with the last question
First remove the brackets, and then merge the similar items: Question 1 - (2x-3y) + (- x + 2Y) question 2 the square of a (the square of ab-b) - the square of B question 3 (the square of x)
Question 1 - (2x-3y) + (- x + 2Y)
The second problem is the square of A-2 (the square of ab-b) - the square of B
The third problem (the square of X - the square of Y) - 3 (the square of 2x - the square of 3Y)
-(2x-3y)+(-x+2y)=-2x+3y-x+2y=-3x+5ya²-2(ab-b²)-b²=a²-2ab+2b²-b²=a²-2ab+b²=(a-b)²(x²-y²)-3(2x²-3y²)=x²-y²-6x²+9y²=-5...
Factorization of 3 (x-1) ^ 3 y - (1-x) ^ 3 Z
Children's shoes, you see the formula, put forward the negative sign of - x in the second half of - (1-x) ^ 3Z, and it becomes 3 (x-1) ^ 3 + (x-1) ^ 3Z. Then you can put forward the common (x-1) ^ 3, so 3 (x-1) ^ 3 y - (1-x) ^ 3Z factorization = (x-1) ^ 3 (3Y + Z)
3(x-1)^3 y-(1-x)^3 z =（x-1)^3（3y+z）