Given a = {x | x ^ 2-3x 2 = 0, X ∈ r}, B = {x | 0 ＜ x ＜ 5. X ∈ n}, then a contains What is the number of sets C contained in B?

Given a = {x | x ^ 2-3x 2 = 0, X ∈ r}, B = {x | 0 ＜ x ＜ 5. X ∈ n}, then a contains What is the number of sets C contained in B?

A={x|x^2-3x+2=0,x∈R}={1,2}
B=｛x|0＜x＜5.x∈N｝={1,2,3,4}
Satisfy the condition that a is included and C is included in B
A is a subset of C and C is a subset of B,
Then C must contain 1,2
C={1,2}
C={1,2,3}
C={1,2,4}
C={1,2,3,4}
There are four sets C satisfying the conditions
Let P = {x | X & # 178; - 2 √ 3x ≤ 0}, what is the relationship between {m} of degree 0.3 of M = 2 and P? The answer is that {m} is the proper subset of P
P={x|x²-2√3x≤0}={x|0≤x≤2√3}
M = 0.3 power of 2, greater than 0 power of 2, less than 1 power of 2, namely 1
If 2A = 3B = 4C, then a: B: C=______ .
Let 2A = 3B = 4C = k, a = K2, B = K3, C = K4, a: B: C = K2: K3: K4 = 6:4:3
4ax^3y-2ax^2y^2-4axy+2ay^2
(x^2-3x-8)(x^2-3x-6)-8
1) =4axy(x^2-1)-2ay^2(x^2-1)=(x-1)(x+)(4axy-2ay^2)=2ay(x-1)(x+1)(2x-y)2) =[(x^2-3x-6)-2](x^2-3x-6)-8=(x2-3x-6)^2-2(x^2-3x-6)-8=[(x2-3x-6)-4][(x2-3x-6)+2]=(x-5)(x+2)(x-4)(x+1)
(1) Original formula = 4ax ^ 3y-4ax-2ax ^ 2Y ^ 2 + 2ay ^ 2
=4axy(x^2-1)-2ay^2(x^2-1)
=(x^2-1)(4axy-2ay^2)
=2ay(x+1)(x-1)(2x-y)
(2) (x^2-3x-8)(x^2-3x-6)-8
=[(x ^...)
(1) Original formula = 4ax ^ 3y-4ax-2ax ^ 2Y ^ 2 + 2ay ^ 2
=4axy(x^2-1)-2ay^2(x^2-1)
=(x^2-1)(4axy-2ay^2)
=2ay(x+1)(x-1)(2x-y)
(2) (x^2-3x-8)(x^2-3x-6)-8
=[(x^2-3x-6)-2](x^2-3x-6)-8
=(x^2-3x-6)^2-2(x^2-3x-6)-8
=[(x^2-3x-6)-4][(x^2-3x-6)+2]
=(x^2-3x-10)(x^2-3x-4)
=(X-5) (x + 2) (x-4) (x + 1) fold up
If 2a-b = 2 is known, then - 2A + B = (), - 4A + 2b-5 = () the negative eighth of the square of 2008 × the negative eighth of the square of 2009
If 2a-b = 2, then - 2A + B = (- 2), - 4A + 2b-5 = (- 9)
Minus one eighth of the square of 2008 × minus eight of the square of 2009
=2008 power of (- 1 / 8) × (- 8)
=2008 power of [(- 1 / 8) * (- 8)] × (- 8)
=2008 power of 1 × (- 8)
=-8
Factorization two, tomorrow to hand. Gods help me!
1-4x^2+4xy-y^2 x^2+2x+1-a^2-2ab-b^2
1. The original formula = 1 - (x-2y) &# = (1 + x-2y) (1-x + 2Y)
2. The original formula = (x + 1) & #178; - (a + b) & #178; = (x + 1 + A + b) (x + 1-a-b)
1-(4x^2-4xy+y^2)=1-(2x-y)^2=(1-2x+y)(1+2x-y)
(x+1)^2-(a+b)^2=(x+1-a-b)(x+1+a+b)
①－1－（ 2X+Y） ²
II.
If the square of 4A - the square of B = 42,2a + B = 14, then 2b-4a =?
2a-b
=(square of 4A - square of B) / (2a + b)
=42÷14
=3
2b-4a
=-2*3
=-6
（2a+b）（2a-b）=42
So 2a-b = 3
2b-4a=-2（2a-b）=-6
X squared minus 2x minus 3 2x squared minus x minus 1 x squared plus 3x minus 4 3x squared minus 8x minus 3
Gods, please!
x^2-2x-3=(x+1)(x-3)
2x^2-x-1=(2x+1)(x-1)
x^2+3x-4=(x-1)(x+4)
3x^2-8x-3=(3x+1)(x-3)
Be sure to answer this question before 13:00 on July 22. The math experts know that 4A ^ 2 + B ^ 2 + 12a-8b + 25 = 0 (2a-b) (a + 2b)
4a²+b²+12a-8b+25=0
4(a²+3a+9/4)+(b²-8b+16)=0
4(a+3/2)²+(b-4)²=0
The solution is a = - 3 / 2, B = 4
(2a-b)(a+2b)=-7×13/2=-91/2=-45.5
The above formula can be changed into (2a + 3) ^ 2 + (B-4) ^ 2 = 0, so a = - 1.5, B = 4, which is (- 3-4) (- 1.5 + 8) = - 45.5
2x-4y-xy+2y^2
ab+ac-b^2-bc
4a^2+4ab+b^2-1
c^2-a^2-2ab+1
x^2-4y^2+12yz-9z^2
a^2b^2-c^2+2ab+1
c^2-a^2-2ab+b^2
（x+y）（2-y）
（a-b）（b+c)
（2a+b）（2a-b）
4. Question 5 is wrong
（ab+1+c）（ab+1-c）