Given a = {x | x ＜ a}, B = {x | x2-3x + 2 ＜ 0} and a ∪ (∁ RB) = R, then the value range of real number a is () A. a≤1B. a＜1C. a≥2D. a＞2

Given a = {x | x ＜ a}, B = {x | x2-3x + 2 ＜ 0} and a ∪ (∁ RB) = R, then the value range of real number a is () A. a≤1B. a＜1C. a≥2D. a＞2

By B = {x | x2-3x + 2 ＜ 0} = {x | 1 ＜ x ＜ 2}, so ∁ RB = {x | x ≤ 1 or X ≥ 2}, and a = {x | x ＜ a}, and a ∪ (∁ RB) = R, so a ≥ 2
Given a = {x | x2-3x + 2 = 0}, B = {x | ax2-2ax + a-5 = 0} if a ∩ B = B, find the value range of A
Given a = {x | x2-3x + 2 = 0}, B = {x | ax2-2ax + a-5 = 0} if a ∩ B = B, find the value range of A
A={X|X^2-3X+2=0}={1,2},A∩B=B
1.B=φ→(-2a)^2-4a(a-5)
One
A=5
1. It is proved that the product of four continuous natural numbers plus one must be a complete square number
2. Calculation: the quotient of the third power of 2003 minus 2 times the second power of 2003 minus 2001 divided by the third power of 2003 plus the second power of 2003 minus 2004
Namely: (2003 ` 3-2 * 2003 ` 2-2001) / (2003 ` 3 + 2003 ` 2-2004)
3. The perimeter of a rectangle is 16cm, its two sides X and y are positive integers, and satisfy x-y-x ` 2 + 2xy-y ` 2 + 2 = 0, then calculate its area
1.(n-2)*(n-1)*n*(n+1)+1=n^4-2n^3-n^2+2n+1=n^4-2n^2(n+1)+(n+1)^2=[n^2-(n+1)]^2
2. Let x = 2003, then 2001 = x-22004 = x + 1
Then (x ^ 3-2x ^ 2-x + 2) / (x ^ 3 + x ^ 2-x-1) = [(x ^ 2-1) (X-2)] / [(x ^ 2-1) (x + 1)] = (X-2) / (x + 1) = 2001 / 2004
3. X + y = 8, x = 8-y, then (8-2y) - (8-2y) ^ 2 + 2 = 0, then [(8-2y) - 2] * [(8-2y) + 1] = 0
So 8-2y = 2 or - 1 (rounding off), y = 3, x = 5
B2-4a2-4a-1 = factorization
The original formula = the square of B = (B + 2A + 1) (b-2a-1)
b²-4a²-4a-1
=b²-(4a²+4a+1)
=b²-(2a+1)²
=[b+(2a+1)][b-(2a+1)]
=(b+2a+1)(b-2a-1)
M + 1 power of (a + x) × (B + x) n-1 power minus m power of (a + x) × (B + x) n power
Factorization
The original formula (a + x) m + 1 power × (B + x) n-1 power minus (a + x) m power × (B + x) n power = (a + x) ^ m * (a + x) * (B + x) ^ (n-1) - (a + x) ^ m * (B + x) ^ (n-1) * (B + x) = (a + x) ^ m * (B + x) ^ (n-1) * (a + x-b-x) = (a + x) ^ m * (B + x) ^ (n-1) * (A-B
Decomposition factor: (1) 4a2x2-16x2y2 (2) A2 (A-3) - A + 3 (3) (A2 + 1) 2-4a (A2 + 1) + 4a2 (4) x4-9x2 + 20
（1）4a2x2-16x2y2=4x2（a2-4y2）=4x2（a+2y）（a-2y）；（2）a2（a-3）-a+3＝a2(a−3)−(a−3)＝(a−3)(a2−1)＝(a−3)(a+1)(a−1)（3）（a2+1）2-4a（a2+1）+4a2=（a2+1-2a）2=（a-1）4；（4）x4-9x2+20＝(x2−4)(...
①4(x+y)²-9(x-y)²
②(x²+1)²-20(x²+1)+100
③x²+5x-6
④(a²+a)-10(a²+a)-24
⑤xy-x-y+1
⑥a²-b²-2a+1
①4(x+y)²-9(x-y)²=[2(x+y)+3(x-y)][2(x+y)-3(x-y)]=(5x-y)(-x+5y)②(x²+1)²-20(x²+1)+100=(x²+1-10)²=(x²-9)²=(x+3)²(x-3)²;③x²+5x-6=(x+6)(x-1);④...
The fourth power of a, the N + 2 power of B and the M-1 power of 5A, the 2n-3 power of B are the same kind of terms, so we can find the value of (M + n) (m-n)
From the meaning of the title: M-1 = 4, N + 2 = 2n-3
We get: M = 5, n = 5
So: (M + n) (m-n) = 0
Zero
——————————————————————】
①4x²-5
②2x²+3x-1
③4x²+4xy-y²
④4x²y²+4xy-1
①4x²-5=(2x-√5)(2x+√5)
② Let 2x & sup2; + 3x-1 = 0
△=3²-4*2*(-1)=17
x1=(-3+√17)/4 x2=(-3-√17)/4
Ψ 2x & sup2; + 3x-1 = 2 * (x-x1) * (x-x2) (substituting X1 and x2)
③ Let 4x & sup2; + 4xy-y & sup2; = 0 and 4x & sup2; Y & sup2; + 4xy-1
Take x as the unknown number to solve the equation
①4x²-5=(2x-√5)(2x+√5)
② Let 2x & sup2; + 3x-1 = 0
△=3²-4*2*(-1)=17
x1=(-3+√17)/4 x2=(-3-√17)/4
Ψ 2x & sup2; + 3x-1 = 2 * (x-x1) * (x-x2) (substituting X1 and x2)
③ Let 4x & sup2; + 4xy-y & sup2; = 0 and 4x & sup2; Y & sup2; + 4xy-1
1.(2x-√5)(2x+√5)
3.（2x-y）²
Directly.. No process
2.4 two questions.. I'm too lazy to do it
Originally, I wanted to copy it in the children's shoes above
It turns out he's not factoring..
If AmBn + 6 is the same as 5a2n + 1b3m + 3, then the value of m-2005n is______ .
From the definition of similar terms, we can know: M = 2n + 13m + 3 = n + 6, solution, M = 1, n = 0, then m-2005n = 1