We know two sets a = {x | x ^ 2-3x + 2 = 0}, B = {x | x ^ 2-ax + A-1 = 0} Is there a real number a so that B is not a subset of a? If so, find the set of all the values of A. if not, explain the reason

We know two sets a = {x | x ^ 2-3x + 2 = 0}, B = {x | x ^ 2-ax + A-1 = 0} Is there a real number a so that B is not a subset of a? If so, find the set of all the values of A. if not, explain the reason

A={1,2},
Let a exist so that B is not a subset of A
Then B has a solution and the root is not {1,2}
Then a ^ 2-4 (A-1) = a ^ 2-4a + 4 = (A-2) ^ 2 > = 0
B = {1} does not hold when a = 2
And 1 + 2 ≠ a or 1 * 2 ≠ A-1
Then a ≠ 3
Then the set of all the values of a is {a | a ≠ 3 and a ≠ 2}
existence
A={1，2}
B={1，a-1}
a-1！ =2
a！ =3
A={1,2}
x^2-ax+a-1=0
[x-(a-1)](x-1)=0
x1=a-1 x2=1
If B is not a's own
Then 1, A-1 is not equal to 1, that is, a is not equal to 2
2. A-1 is not equal to 2, that is, a is not equal to 3
In conclusion, a is not equal to 2 or 3
Given u = R, a = {x | x > 0}, B = {x | x ≤ 1}, then (a ∩ cub) ∪ (B ∩ CUA)=_________
∵U=R,A={x| x＞0},B={x|x≤1}
∴CuB=｛x│x＞1｝
∴A∩CuB={x| x＞0}
Also ∵ CUA = {x │ x ≤ 0}
∴B∩CuA={x|x≤1}
(a ∩ cub) ∪ (B ∩ CUA) = {x │ x ＞ 0 or X ≤ 1}
Let u = R, a = {x | 0 ＜ x ＜ 5}, B = {x | x ＞ - 1} find CUA and cub
What is the square of (2a-b) - 4 (a + b) (a-2b) equal to
Square of (2a-b) - 4 (a + b) (a-2b)
=4a^2-4ab+b^2-4(a^2-ab-2b^2)
=4a^2-4ab+b^2-4a^2+4ab+8b^2
=9b^2
Square of (2a-b) - 4 (a + b) (a-2b)
=4a^2-4ab+b^2-4a^2+4ab+8b^2
=9b^2
(2a-b)²-4(a+b)(a-2b)
=(2a-b)[(2a-b)-4(a+b)]
=(2a-b)(2a-b-4a-4b)
=-(2a-b)(2a+5b)
The process... It's a long time
=4a-4ab + B - (4a-8ab + 4ab-8b)
=4a-4ab + b-4a + 4AB + 8B
=9b
Calculation by factorization
2001 m3 - 2 * 2001 M2 - 1999 / 2001 m3 + 2001 M2 - 2002
=(X(X^2-1)-2(X^2-1))/(X(X^2-1)+X^2-1)
=(X-2)/(X+1)
I don't quite understand this step
2001 m3 - 2 * 2001 M2 - 1999 / 2001 m3 + 2001 M2 - 2002
=[2001^2(2001-2)-1999]/[2001^2(2001+1)-2002]
=[2001^2*1999-1999]/[2001^2002-2002]
=[1999(2001^2-1)]/[2002(2001^2-1)]
=1999/2002
Hello!
Let x = 2001
(X^3-2X^2-X+2)/(X^3+X^2-X-1)
=(X(X^2-1)-2(X^2-1))/(X(X^2-1)+X^2-1)
=(X-2)/(X+1)
=1999/2002
A + B + (the square of 2B / 2a-b) is equal to less
The best solution out, I am a junior two students, the best let me see!
a+b+2b^2/(2a-b)=[(a+b)(2a-b)+2b^2]/(2a-b)=(2a^2+ab+b^2)/(2a-b)
In this way, you can't factorize and simplify after general division. If you don't believe in your own checking, the numerator and denominator are irreducible!
Original formula = (2a ^ 2-AB + 2ab-b ^ 2 + 2B ^ 2) / 2a-b -- General Division
=(2a ^ 2 + ab-b ^ 2) / 2a-b = (2a-b) (a + b) / 2a-b -- factorization
=a+b
x^2 y - 2 x y = ________ ;
1 - x + 1/4 x^2 = __________ ;
a^4 - 2a^2 + 1 =_______ ;
（ a - b ）^2 - 4 =_______ .
x^2 y - 2 x y =xy（x-2）
1 - x + 1/4 x^2 =（1-1/2x）^2
a^4 - 2a^2 + 1 =(a^2-1)^2
（ a - b ）^2 - 4 =(a-b+2)(a-b-2)
x²y-2xy=xy(x-2)
1-x+1/4x²=1²-2×1×(1/2x)+(1/2x)²=(1-1/2x)²
a^4-2a²+1=(a²)²-2×a²+1²=(a²-1)²=[(a+1)(a-1)]²=(a+1)²(a-1)²
(a-b)²-4=[(a-b)+2][(a-b)-2]=(a-b+2)(a-b-2)
When A-B of a + B = 3, the value of square-5 of 2A + 2B + (a-b of a + b) is?
A-B 2A + 2B + (a + B a-b) = twice the square of a + B + (a + B a-b) = 2 / 3 + 3 = 2 / 3 + 9, and then subtract 5 to get 14 / 3
The eighth grade first volume mathematics, uses the factor decomposition computation to obtain the result
Given 6x & # 178; - 13xy + 6y & # 178; = 0, find the value of X / y
(3x-2y)(2x-3y)=0
X / y = 2 / 3 or 3 / 2
Given the square of a = 2A - A, B = - 5 + A + 1, find the value of 3a-2b + 1 when a = 1 / 2
The answer is 8