Given the complete set I = R, set a = {x | x ^ 2 + 3x + 2 = 0}, B = {x | x ^ 2 + (M + 1) x + M = 0}. If CIA intersection B = Φ, what is the process of finding the value of M? Is the value of m not equal to 2?

Given the complete set I = R, set a = {x | x ^ 2 + 3x + 2 = 0}, B = {x | x ^ 2 + (M + 1) x + M = 0}. If CIA intersection B = Φ, what is the process of finding the value of M? Is the value of m not equal to 2?

The value of M is equal to 2. Set a = {- 1, - 2}, so CIA does not include - 1, - 2. Because CIA intersection B = Φ, so B is a subset of A. There are three cases: 1) if B is an empty set, then (M + 1) ^ 2-4m = (m-1) ^ 2 less than 0 does not hold; 2) if B contains one element, then (m-1) ^ 2 is equal to 0, M = 1, then x = - 1 holds; 3) if B contains two elements
A={-1,-2}
CIA = {x | X-1 and X-2}
Because CIA pays B = Φ
The discussion is divided into three situations
1. B=Φ
(m+1)^2-4*m
The definition field of function f (x) = √ (X & # 178; - 2ax-a) is r, and the value range of real number a is obtained
The domain of F (x) = √ (X & # 178; - 2ax-a) is r
That is X & # 178; - 2ax-a ≥ 0
∵ the quadratic coefficient is positive
The discriminant = 4A & # 178; + 4A ≤ 0
The value range of real number a is - 1 ≤ a ≤ 0
B ^ 2-4ac
Integral multiplication and division and factorization,
Note: those drawing horizontal lines are superscripts
Given a & sup2; N + 1 = 5. Find the value of a 6N + & sup3
—— ——
6n+3=3(2n+1)
So the (6N + 3) power of A
=[2n + 1 power of a] & sup3;
=5³
=125
Given 4a2 + 4A + b2-6b + 10 = 0, find the value of a3b + AB3
4a²+4a+b²-6b+10=0
(4a²+4a+1）+（b²-6b+9）=0
(2a+1)²+（b-3）²=0
2a+1=0
b-3=0
a=-0.5
B=3
a3b+ab3
=ab(a2+b2)
=-1.5*(0.25+9)
=-13.875
4a2+4a+b2—6b+10=0
（4a²+4a+1)+(b²-6b+9)=0
(2a+1)²+(b-3)²=0
2a+1=0,b-3=0
a=-1/2, b=3
a³b+ab³=ab(a²+b²)
=-3/2×﹙¼+9)
=-111/8
(2a^2+4a+1^2)+(b^2-6b+3^2)=0 a^3b+ab^3=(-0.5)^3*3+(-0.5)*3^3
=-0.125 *. Deployment
(2a^2+4a+1^2)+(b^2-6b+3^2)=0 a^3b+ab^3=(-0.5)^3*3+(-0.5)*3^3
=-0.125*3+(-0.5)*27
=-0.375-13.5
=-13.875
Using the complete square formula
(2a+1)^2+(b-3)^2=0
Only 0 + 0 or the opposite numbers add up to 0, but they are all square, so they can't be the opposite numbers
2a+1=0,b-3=0
2a=-1,b=3
A = - 0.5 ﹣ put away
Help me solve a factoring problem in grade two
Factorization: (a + b) ^ n + 2-2 (a + b) ^ n + 1 + (a + b) ^ n
(a+b)^(n+2)-2(a+b)^(n+1)+(a+b)^n
=(a+b)^n*【(a+b)^2-2(a+b)+1】
==(a+b)^n * (a+b-1)^2
Given a 2 + B 2-4a-6b + 13 = 0, find the value of a-b
∵ A2 + b2-4a-6b + 13 = a2-4a + 4 + b2-6b + 9 = (A-2) 2 + (B-3) 2 = 0, ∵ A-2 = 0, B-3 = 0, i.e. a = 2, B = 3, then A-B = 2-3 = - 1
1. Factorization factor (x + 2) (x + 4) + x ^ 2-4
2. Simplify first and then evaluate: [(X-Y) ^ 2 + (x + y) (X-Y)] / 2x, where x = 3, y = 1.5
3. Factorization - 4 ^ 2Y + 4xy ^ 2-y ^ 3
-4X ^ y + 4xy ^ 2-y ^ 3... Wrong third question
1.=x^2+6x+8+x^2-4=2x^2+6x+4=2(x^2+3x+4)=2(x+1)(x+2)
2.=(x^2+y^2-2xy+x^2-y^2)/2=x-x=1.5
Is the title wrong?
1.)（x+2)(x+4)+x^2-4
=2x^2+6x+4
=(2x+2)(x+2)
2)[(x-y)^2+(x+y)(x-y)]÷2x
=[x^2-2xy+y^2+(x^2-y^2)]/2x
=[x^2-2xy+x^2]/2x
=x-y
=1.5
2) Original formula = [(X-Y) (X-Y + X + y)] / 2x
=[2X(X-Y)]÷2X
=X-Y
When x = 3, y = 1.5
Original formula = 3-1.5
=1.5
1、（x+2)(x+4)+x^2-4.
=x^2+4x+2x+8+x^2-4
=2x^2+6x+4
=2(x^2+3x+2)
=2(x+1)(x+2)
2、[(x-y)^2+(x+y)(x-y)]÷2x
=[x^2+y^2-2xy+x^2-y^2]÷2x
=[2x^2-2xy]÷2x
=x-y
When x = 3, y = 1.5
x-y=1.5
3、-4x^2y+4xy^2-y^3
=-y(4x^2-4xy+y^2）
=-y(2x-y)^2
(1) Original formula = x ^ 2 + 4x + 2x + 8 + x ^ 2-4
=2x^2+6x+4
=2(x^2+3x+2)
=2[x(x+3)+2]
(2) Original formula = [x ^ 2-2xy + y ^ 2 + x ^ 2-y ^ 2] / 2x
=[2x^2-2xy]÷2x
=x-y
... unfold
(1) Original formula = x ^ 2 + 4x + 2x + 8 + x ^ 2-4
=2x^2+6x+4
=2(x^2+3x+2)
=2[x(x+3)+2]
(2) Original formula = [x ^ 2-2xy + y ^ 2 + x ^ 2-y ^ 2] / 2x
=[2x^2-2xy]÷2x
=x-y
Substitute x = 3, y = 1.5 to get
Original formula = 3-1.5
=1.5
(3) Original formula = y (- 4 ^ 2 + 4xy-y ^ 2)
=-y(4^2-4xy+y^2)
=-y(4^2-y^2+2y^2-4xy)
=-y[(4+y)(4-y)+2y(y-2x)]
I'm also a student of grade two in junior high school. Maybe I'm not all right. Put it away
Question 1: (x + 2) (x + 4) + x ^ 2-4 = x ^ 2 + 6x + 8 + x ^ 2-4 = 2x ^ 2 + 6x + 4 = 2 (x ^ 2 + 3x + 2) = 2 (x + 1) (x + 2)
The second question: first expand the collation, and then into the evaluation, as follows
(x^2-2xy+y^2+x^2-y^2)/2x=2(x^2-xy)/2x=x-y=3-1.5=1.5
Question 3: are you sure your question is correct? Only one X can't be decomposed. Please reconfirm.
a2＋b2＋4a＋6b＋13=0
a. B is a real number, find ab
Help solve ~ factorization
Title: (a + b) ^ 2-8 (a + b) + 16
Please write down the whole process and explanation for me~
(a+b)^2-8(a+b)+16
=(a+b-4)^2
Is the complete square formula. The square of a + B-4
Perfect square!
=(a+b-4)^2
Complete square difference formula:
Original formula = (a + B-4) ^ 2
Friends upstairs: not + 4, but - 4!! Or where did it come from - 8
For tile, (a + B-4) ^ 2
The one on the first floor is a bit stupid
Factorization of 4a-4a ^ 2-1
4a-4a^2-1=(1-2a)(2a-1)=-(2a-1)^2
Cross method
-4a^2+4a-1
-2 1
2 -1
(-2)*(-1)+2*1=4
Associative complete square formula
-4a^2+4a-1=-(4a^2-4a+1)=-((2a)^2-2*(2a)+1)
Undetermined coefficient
-4a^2+4a-1=(ax-1)(bx+1)=abx+(a-b)x-1
ab=-4,a-b=4
a=b+4
(b+4)b=-4
b^2+4b+4=0
b=-2,a=2
Or try to figure out B = - 2, a = 2
-4*（a-1/2）^2