Given the set a = {x | x2-3x + 2 = 0}, B = {x | x2-mx + 2 = 0}, if a is a necessary and insufficient condition of B, the range of real number m is obtained

Given the set a = {x | x2-3x + 2 = 0}, B = {x | x2-mx + 2 = 0}, if a is a necessary and insufficient condition of B, the range of real number m is obtained

A ∩ B = B ⇔ B ⊆ A is a necessary and insufficient condition of B. according to the number of elements in the set B, B = ∈, B = {1} or {2}, B = {1,2} when B = ∈, △ = m2-8 ∩ 0 ∩ 22 ∩ m < 22; when B = {1} or {2}, △ = 01-m + 2 = 0 or 4-2m + 2 = 0, M has no solution; when B = {
The known set a = {x | - x ^ 2 + 3x + 10 = > 0}, B = {x | K + 1}
Set a = {x | - x ^ 2 + 3x + 10 = > 0}
-x^2+3x+10=>0
We get (- x + 5) (x + 2) = > 0, that is, two factors should be = > 0 or 0, x + 2 = > 0 or - x + 5 at the same time
(x-m) (x-1) = x & # 178; - 3x + N, then the value of M + n is?
(x-m)(x-1)=x²-（1+m）x+m=x²-3x+n
Then, 1 + M = 3, M = 2; m = n = 2
So, M + n = 2 + 2 = 4
m=n=2, m+n=4
(x-m)(x-1)=x²-mx-x+m==x²-(m+1)x+1=x²-3x+n
It is known that (M + 1) = 3, M = n;
m=n=2
m+n=4
M (x + 2) / (x + 1) (x + 2) - n (x-1) / (x + 1) (x + 2) = 1 / X & # 178; + 3x + 2 m, n is not 0 constant, find 1 / (M + n) &# 178;
(x+1)(x+2)=x^2+3x+2
If the denominators on both sides are the same, just compare the coefficients of the numerator
The numerator on the left is
mx+2m-(nx-n)
=(m-n)x+(2m+n)
The right molecule is 1
therefore
m-n=0
2m+n=1
The solution is m = 1 / 3, n = 1 / 3
1/(m+n)^2=9/4
If you still have questions, please ask!
1）y^2-3y-2
2)-6x^2+2x+1
3)x^4-13x^2+36
4)5x^3y-2x^y^2-xy^3
1) [y-(3+√17)/2][y-(3-√17)/2]
3) (X^2-4)(X^2-9)=(X-2)(X+2)(X-3)(X+3)
Like 5 to the 23rd power - 5 to the 21st power can be divisible by 120
5^23-5^21=5^21*5^2-5^21=(5^2-1)*5^21=24*5^21=24*5*5^20=120*5^20
So it's divisible by 120
The first day of junior high school uses the method of grouping and factoring!
There is a process before the answer
Square of X + 2XY + square of Y-1=
The square of a, the square of B - the square of a + 2Ab + 1=
The square of x-x-9y-3y=
Square of x-y-z-2yz=
It is known that a = (x + 2) (x-3) (x + 4) (X-5) + 49
Proof: A is a complete square number
It is known that the square of 4x + 4xy + y-4x-2y + 1 = 0
Prove: 2x square + 3xy + y square - X-Y = 0
The third power of X + the third power of X & sup2; y-xy & sup2; - y=
ax²-bx²+bx-ax+a-b=
X & sup2; + 6xy + 9y & sup2; - 16A square + 8a-1=
The square of a-6ab + 12b + 9b & sup2; - 4A=
The fourth power of a - the third power of 2A + A & sup2; - 9=
4a²x-4a²y-b²x+b²y=
x²-2xy-xz+yz+y²=
Square of X + 2XY + square of Y-1 = (x + y) ^ 2-1 = (x + y + 1) (x + Y-1)
Square of a square of b-square of a + 2Ab + 1 = (AB + 1) ^ 2-A ^ 2 = (AB + A + 1) (AB-A + 1)
The square of X - the square of x-9y - 3Y = (x + 3Y) (x-3y) - (x + 3Y) = (x-3y-1) (x + 3Y)
Square of X - square of Y - square of Z - 2yz = (Y-Z) ^ 2-x ^ 2 = (Y-Z + x) (y-z-x)
It is known that a = (x + 2) (x-3) (x + 4) (X-5) + 49
Proof: A is a complete square number
A=(x^2-x-6)(x^2-x-20)+49
=(x^2-x)^2-26(x^2-x)+120+49
=(x^2-x)^2-26(x^2-x)+169
=(x^2-x-13)^2
It is known that the square of 4x + 4xy + y-4x-2y + 1 = 0
Prove: 2x square + 3xy + y square - X-Y = 0
4x^2+4xy+y^2-4x-2y+1=0
(2x+y)^2-2(2x+y)+1=0
(2x+y-1)^2=0
2x+y-1=0
2x^2+3xy+y^2-x-y
=(2x+y)(x+y)-(x+y)
=(2x+y-1)(x+y)
=0
The third power of X + the third power of X & sup2; y-xy & sup2; - y
=x^2(x+y)-y^2(x+y)
=(x^2-y^2)(x+y)
=(x+y)(x-y)(x+y)
=(x-y)(x+y)^2
ax²-bx²+bx-ax+a-b=x^2(a-b)-x(a-b)+(a-b)=(x^2-x+1)(a-b)
X & sup2; + 6xy + 9y & sup2; - 16A square + 8a-1
=(x+3y)^2-(4a-1)^2
=(x+3y+4a-1)(x+3y-4a+1)
The square of a-6ab + 12b + 9b & sup2; - 4A = (a-3b) ^ 2-4 (a-3b) = (a-3b-4) (a-3b)
The fourth power of a - the third power of 2A + A & sup2; - 9 = (a ^ 2-A) ^ 2-9 = (a ^ 2-A + 3) (a ^ 2-a-3)
4a²x-4a²y-b²x+b²y
=4a^2(x-y)-b^2(x-y)
=(4a^2-b^2)(x-y)
=(2a+b)(2a-b)(x-y)
x²-2xy-xz+yz+y²=(x-y)^2-z(x-y)=(x-y-z)(x-y)
1. The original formula = (x ^ 2 + 2XY + y ^ 2) - 1
=（x+y）^2-1^2
=(x+y+1)(x+y-1)
2. The original formula = (a ^ 2B ^ 2 + 2Ab + 1) - A ^ 2
=(ab+1)^2-(a^2)
=(ab+1+a)(ab+1-a)
3. The original formula = (x ^ 2-9y ^ 2) + (x-3y)
=(x + 3Y) (x-3... Expansion
1. The original formula = (x ^ 2 + 2XY + y ^ 2) - 1
=（x+y）^2-1^2
=(x+y+1)(x+y-1)
2. The original formula = (a ^ 2B ^ 2 + 2Ab + 1) - A ^ 2
=(ab+1)^2-(a^2)
=(ab+1+a)(ab+1-a)
3. The original formula = (x ^ 2-9y ^ 2) + (x-3y)
=(x+3y)(x-3y)+(x-3y)
=(x-3y)(x+3y+1)
4. The original formula = x ^ 2 - (y ^ 2 + 2yz + Z ^ 2)
=x^2-(y+z)^2
=(x-y-z)(x+y+z)
5. The original formula = (x ^ 2-x-6) (x ^ 2-x-20) + 49
Let x ^ 2-x-7 = a, then the original formula = (x + 7) (X-7) + 49
=x^2-49+49
=x^2
So a is a perfect square.
6.4x square + 4xy + y square - 4x-2y + 1 = (2x + y) ^ 2-2 (2x + y) + 1
=(2x+y-1)^2=0
So 2x + Y-1 = 0
So the square of 2x + 3xy + Y - X-Y = (2x + y) (x + y) - (x + y)
=(2x+y-1)(x+y)
=0
7. Original formula = (x ^ 3 + x ^ 2Y) - (y ^ 3 + XY ^ 2)
=(x^2-y^2)(x+y)
=(x+y)(x-y)(x+y)
=(x+y)^2(x-y)
8. The original formula = x ^ 2 (a-b) - x (a-b) + (a-b)
=(x^2-x+1)(a-b)
9. The original formula = (x + 3Y) ^ 2 - (4a-1) ^ 2
=(x+3y-4a+1)(x+3y+4y-1)
10. The original formula = (a ^ 2-6ab + 9b ^ 2) + (12b-4a)
=(3b-a)^2+3(3b-a)
=(3b-a+3)(3b-a)
11. The original formula = a ^ 2 (a ^ 2-2a + 1) - 3 ^ 2
=(a^2-a)^2-3^2
=(a^2-a-3)(a^2-a+3)
12. Original formula = 4A ^ 2 (X-Y) - B ^ 2 (X-Y)
=(4a^2-b^2)(x-y)
=(2a+b)(2a-b)(x-y)
13. The original formula = (x ^ 2-2xy + y ^ 2) - (XZ YZ)
=(x+y)^2-z(x+y)
=(x+y-z)(x+y)
Warm tips: group decomposition, mainly using the idea of formula and the whole idea, but also pay attention to the change of symbols. come on. Put it away
Can the 23rd power of 5 - the 21st power of 5 be divisible by 120
Factorization problem algebra is best to use a simple group decomposition method, but some can not be used do not use
（1）x²-4x+4-y² (2)c²-a²-b²+2ab （3）x²_ 2xy+y²-ab+ay (4)a²-4ab+4b²-6ab²+3a²b
The third one is wrong, which is X & # 178; - 2XY + Y & # 178; - ax + ay
That step, the group step, it's better to write it out
Do two triangles are congruent? Please give an example and explain the reason
It's not necessary to make an isosceles triangle. It's not necessary to be congruent to find a point at the bottom of an isosceles triangle connecting with the vertex