Given the complete set u = {1,2,3,4}, a = {x | X & # 178; - 5x + M = 0, X ∈ u}, find the complement of a and m

Given the complete set u = {1,2,3,4}, a = {x | X & # 178; - 5x + M = 0, X ∈ u}, find the complement of a and m

If x ∈ u, then
(1) When x = 1, 1 & # 178; - 5 + M = 0, the solution is m = 4
In this case, the equation is X & # 178; - 5x + 4 = 0, and the solution is x = 1 or 4
In this case, a = {1,4}
therefore
The complement of a is {2,3} M = 4
(2) When x = 2, 2 & # 178; - 5 * 2 + M = 0, the solution is m = 6
In this case, the equation is X & # 178; - 5x + 6 = 0, and the solution is x = 2 or 3
In this case, a = {2,3}
therefore
The complement of a is {1,4} M = 6
(3) When x = 3, 3 & # 178; - 5 * 3 + M = 0, the solution is m = 6
In this case, the equation is X & # 178; - 5x + 6 = 0, and the solution is x = 2 or 3
In this case, a = {2,3}
therefore
The complement of a is {1,4} M = 6
(4) When x = 4, 4 & # 178; - 5 * 4 + M = 0, the solution is m = 4
In this case, the equation is X & # 178; - 5x + 4 = 0, and the solution is x = 1 or 4
In this case, a = {1,4}
therefore
The complement of a is {2,3} M = 4
To sum up
When m = 4, the complement of a is {2,3}
When m = 6, the complement of a is {1,4}
We know the complete set u = {- 4, - 3, - 2, - 1,0,1,2,3,4}, and the set a = {- 3, a + 1, a & # 178;}, B=
The complete set u = {- 4, - 3, - 2, - 1,0,1,2,3,4},
Set a = {- 3, a + 1, a & # 178;}, B = {A-3, 2a-1, a & # 178; + 1}, if a ∪ B = {- 3}, find Cu (a ∪ b)
There must be - 3 in B set, then a = negative 1, the result will come out
If a ^ 2 + B ^ 2-4a-8b + 20 = 0, find the value of B ^ - A
a^2-4a+4+b^2-8b+16=0
(a-2)^2+(b-4)^2=0
(a-2)^2≥0
（b-4)≥0
So A-2 = 0. B-4 = 0,
a=2,b=4
b^(-a)=4^(-2)=1/16
（a-2)^2+(b-4)^2=0
a=2 b=4
b^-a=4^-2=1/16
(X & # 178; + x) &# 178; - 26 (X & # 178; + x) + 120 factorization
(x²+x)² -26（x²+x）+120
=(x²+x-6)(x²+x-20)
=(x-2)(x+3)(x-4)(x+5)
20*6=120
(x²+x)² -26（x²+x）+120
=(x²+x-6)(x²+x-20)
=(x+3)(x-2)(x+5)(x-4)
When the value of a / B is, the value of the square of polynomial a-4a 6A 13 is 0
+Or is 6b-6b-6b-6b? To take + 6B as an example: A & #178; + B & \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\+ 9 = 0 (A-2) & # 178; + (B-3) & # 17
（x+y）²-2(x²-y²)+（y-x）²
(x+y）²-2(x²-y²)+（y-x）²
=(x+y）²-2(x+y)(x-y)+（x-y）²
=[(x+y)-(x-y)]²
=(2y)²
=4y²
（x+y）²-2(x²-y²)+（y-x）²=x^2+2xy+y^2-2x^2+2y^2+y^2-2xy+x^2=4y^2
(x + y) square - (X-Y) square-2 (X-Y Square)
=(x + y + X-Y) (x + Y-X + y) - 2 (xsquare-ysquare)
=2X*2Y-2(x²-y²)
=2(2XY-x²+y²)
=2(y-x）²
（x+y）²-2(x²-y²)+（y-x）²
=x²+2xy-2x²+2y²+y²-2xy+y²
=x²+y²-2x²+2y²+y²+y²
=5y²-x²
The original formula = (x + y) & sup2; - 2 (x + y) (X-Y) + (X-Y) & sup2;
=[(x+y)-(x-y)]²
=4y²
Using the complete square formula
The square of 2Y. Do it with the complete square formula
The answer 4Y & sup2; is right, everything else is wrong!
Classmate, you come to the Internet to ask for answers to such a simple question. I hope you can listen to me. Now the social competition is very fierce. In the final analysis, it's the competition for talents. I hope you can study hard, think more about yourself, and don't place your hopes on others. I think if your parents know you're doing this, they must be very sad. Maybe you're only in Junior high school now, and you don't think about life problems. But as a past person, I can tell you clearly that it's very sad that you didn't learn knowledge well It's hard to mix with society. Unless your parents are very rich, very rich, and
The answer 4Y & sup2; is right, everything else is wrong!
Classmate, you come to the Internet to ask for answers to such a simple question. I hope you can listen to me. Now the social competition is very fierce. In the final analysis, it's the competition for talents. I hope you can study hard, think more about yourself, and don't place your hopes on others. I think if your parents know you're doing this, they must be very sad. Maybe you're only in Junior high school now, and you don't think about life problems. But as a past person, I can tell you clearly that it's very sad that you didn't learn knowledge well It's hard to mix with society. Unless your parents have a lot of money, and you're going to depend on them for the rest of your life. Maybe you think I'm nosy and talking nonsense, but I really hope you young people can study hard and understand your parents' hard work. Put it away
Given the square of a - 4A + 9A + 6A + 5 = 0, find the value of one part of a - one part of B
It should be a & sup2; - 4A + 9b & sup2; + 6B + 5 = 0
(a²-4a+4)+(9b²+6b+1)=0
(a-2)²+(3b+1)²=0
If the sum of the squares is 0, it is equal to 0‘
So A-2 = 0, 3b + 1 = 0
a=2,b=-1/3
So 1 / A-1 / b
=1/2+3
=7/2
There is no B in the first formula. How can we get B in the end
It should be a2-4a + 9b2 + 6B + 5 = 0
(a2-4a+4)+(9b2+6b+1)=0
(a-2)2+(3b+1)2=0
If the sum of the squares is 0, it is equal to 0‘
So A-2 = 0, 3b + 1 = 0
a=2,b=-1/3
So 1 / A-1 / b
=1/2+3
=7/2
This is correct, that is to say, dividing 5 points into two sides forms the formula of complete square difference and complete square sum
1.x^2-3x-10
2.x(x+y)(x-y)-x(x+y)^2
3.9(2a+3b)^2-4(3a-2b)^2
4. Given a ^ 2 + A + 1 = 0, find the value of a ^ 3 + 2A ^ 2 + 2A + 3
1.(x-5)(x+2)2.x(x+y)[(x-y)-(x+y)]=-2xy(x+y)3.[3(2a+3b)+2(3a-2b)][3(2a+3b)-2(3a-2b)]=13b(12a+5b)4.a^3+2a^2+2a+3=(a^3+a^2+a)+(a^2+a+1)+2 =a(a^2+a+1)+(a^2+a+1)+2=2
^What symbol?
1. =(x-5)(x+2)
2. =-2xy(x+y)
3. =13b(5b+12a)
4. a^3+2a^2+2a+3=(a^3+a^2+a)+(a^2+a+1)+2
=a(a^2+a+1)+(a^2+a+1)+2=2
Children study hard!!!
If the sum of the square of a and the square of B minus 6A plus 8b plus 25 equals 0, then how much is a and how much is B
aa+bb-6a+8b+25=0
(a-3)(a-3)+(b+4)(b+4)=0
a=3,b=-4
a*a+b*b-6a+8b+25=0
(a-3)(a-3)+(b+4)(b+4)=0
So a = 3, B = - 4
1999^2-1998^2=
-x^2+4xy-4y^2=
4(a-b)^2-9a+9b=
1/2x^2+2xy+2y^2=
1999^2-1998^2=(1999+1998)(1999-1998)=3997 -x^2+4xy-4y^2= -(x^2-4xy+4y^2)=-(x-2y)^24(a-b)^2-9a+9b=4(a-b)^2-9(a-b)=(a-b)(4a-4b-9) 1/2x^2+2xy+2y^2=1/2(x^2+4xy+4y^2)=1/2(x+2y)^2