Let u = {x | 0 ＜ x ＜ 10, X ∈ n *}, if a ∩ B = {3}. A ∩ a ∩ and # 8705; UB = {1,5,7}, CUA ∩ cub = {9}, find a, B

Let u = {x | 0 ＜ x ＜ 10, X ∈ n *}, if a ∩ B = {3}. A ∩ a ∩ and # 8705; UB = {1,5,7}, CUA ∩ cub = {9}, find a, B

A and B = 1.2.3.4.5.6.7.8·
A ∩ &; UB = {1,5,7} so B is not equal to 1,5,7  B = 2,3,4,6,8 because
A ∩ B = {3} so a = 1,3,7
A take. 1，，3，7，。 B take 2, 3, 4, 6, 8..
Let u = R, a = {x | 0}, = x = 1}, find CUA, cub and (CUA) U (cub)
Complete set u = R, a = {x | 0
CUA = {x | x < 0 or x > = 5}
CuB = {x | x < 1 }
(CUA) U (cub) = {x | x < 1 or x > = 5}
There are also examples of the difference between factorization and factorization
Factorization (factorization is actually the same), which transforms a polynomial into the product of several simplest integers. This transformation is called factorization of the polynomial, also called factorization factor. It is widely used in mathematical root drawing
Factorization and factorization are the same thing. It's just called differently. In fact, there is not much difference between these two phrases in the field of mathematics. From a grammatical point of view, there is still a difference: factoring is a verb object
If the square of a + AB + 2B = 0, and B is not equal to 0, find the value of 2a-b of 2A + B, please, today
A ^ 2 + AB + 2B ^ 2 = 0, and B is not equal to 0
Formula (a + 1 / 2 * b) ^ 2 + 7 / 4 * B ^ 2 = 0, so there is a problem with the title
The estimate is: A ^ 2 + ab-2b ^ 2 = 0
Factoring a problem
9（m+n)^2-4(m-n)^2
=(3m+3n-2m+2n)(3m+3n+2m-2n)
=(m+5n)(5m-n)
[3(m+n)]^2-[2(m-n)]^2=(3m+3n+2m-2n)(3m+3n-2m+2n)=(5m+n)(m+5n)
Given a + 2B = 0, find the fraction a ^ 2 + 2ab-b ^ 2 / 2A ^ 2 + AB + B ^ 2
If (x + 1) 2 + (Y-2) 2 = 0, then x2 + Y2=______ If A2 + b2-4a + 2B + 5 = 0, then ab=______ .
(1) Because (x + 1) 2 + (Y-2) 2 = 0, so x = - 1, y = 2, then x2 + y2 = 1 + 4 = 5; (2) because A2 + b2-4a + 2B + 5 = 0, so a2-4a + 4 + B2 + 2B + 1 = 0, that is, (A-2) 2 + (B + 1) 2 = 0, then a = 2, B = - 1. So AB = 2 × (- 1) = - 2
Given A-B = 5, ab = - 2, find (- 2Ab + 2A + 3b) - (3AB + 2b-2a) - (4B + AB)
a-b=5,a=b+5
ab=-2
(b+5)b=-2
b²+5b+2=0
b=[-5±√（5²-4*1*2）]/2=(-5±√17)/2
（-2ab+2a+3b)-(3ab+2b-2a)-(4b+ab)
=4a-3b-6ab
=4a-4b+b-6ab
=4(a-b)-6ab+b
=4*5-6*(-2)+b
=32+b
=32+(-5+√17)/2；
or
The original formula = 32 + (- 5 - √ 17) / 2 = 32 - (5 + √ 17) / 2;
（-2ab+2a+3b)-(3ab+2b-2a)-(4b+ab)
=-6ab+4a-4b
=-6ab+a(a-b)
=-6*(-2)+4*5
=32
（-2ab+2a+3b)-(3ab+2b-2a)-(4b+ab)=32
x^5+2x^4+3x^3-2x^2+x-5
x^5+2x^4+3x^3-2x^2+x-5
=(x^5-x^4)+3(x^4-x^3)+6(x^3-x^2)+4(x^2-x)+5(x-1)
=(x-1)(x^4+3x^3+6x^2+4x+5)
If 1a − 1b = 4, then the value of a − 2Ab − B2A − 2B + 7ab is equal to ()
A. 6B. -6C. 215D. −27
It is known that 1a − 1b = 4 can give A-B = - 4AB, then a − 2Ab − B2A − 2B + 7ab = a − B − 2ab2 (a − b) + 7ab = − 4AB − 2Ab − 8ab + 7ab = − 6ab − AB = 6