The corresponding points of rational numbers a, B and C on the number axis are shown in Figure 2, where o is the origin. B > O > C > A Simplify | A-B | + | a + B | + | B-C | - | a|

The corresponding points of rational numbers a, B and C on the number axis are shown in Figure 2, where o is the origin. B > O > C > A Simplify | A-B | + | a + B | + | B-C | - | a|

B-A+B-A+B-C+A
On the number axis, point a (representing number a) is on the left side of the origin, and point B (representing integer b) is on the right side of the origin. If | A-B | = 2013 and AO = 2Ab, then a plus B =?
A, B, C and D on the number axis are all integers. If the number corresponding to point a is a, the number corresponding to point B is B, b-2a = 7, the origin on the number axis is in a, B, C and D
What and why?
——.——.——.——.——.——.——.——.——
A B C D
C
From the graph, we can see that a is a, B is a + 3, C is a + 4, D is a + 7, B = a + 3, b-2a = 7, the two equations are simultaneous, a = - 4, so C is the origin
Let the sum of the first n terms of the sequence {an} be Sn, and the points (n, Sn / N) (n belongs to n) are all on the image of the function y = 1 / 2x-1 / 2. (2) let BN = 1 / Anan + 1, tn be the sequence {
Let the sum of the first n terms of the sequence {an} be Sn, and the points (n, Sn / N) (n belongs to n) are all on the image of the function y = 1 / 2x-1 / 2,
(2) Let BN = 1 / Anan + 1, tn be the sum of the first n terms of the sequence {BN}, and find TN
Because Sn = n * a1 + n * (n-1) * D / 2
So Sn / N = a1 + (n-1) * D / 2
Because the point is on y = 1 / 2x-1 / 2
So Sn / N = - 1 / 2 + 1 / 2n = 0 + (n-1) * 1 / 2
a1=0 d=1
an=n-1
Then, the formula given by BN is not clear. Can we add brackets where we should? I'll try my best to give you an answer
Sn/n=1/2n-1/2
Sn=1/2-n/2
Sn+1-Sn=-1/2=an+1
When n = 1, A1 = 0
When n = 2, A2 = - 1 / 2
So BN = 5
Tn=5n
(SN) / N = (1 / 2) n-1 / 2, Sn = (1 / 2) n ^ 2 - (1 / 2) n, we know that an is an arithmetic sequence, and an = n-1,
bn=1/（n-1）-1/。 The problem is n = 1, B 1 does not exist, and can be solved only when B 1 is greater than or equal to 2
If Tan α = √ 2, find the value of 2Sin Λ 2 α - sin α cos α + cos Λ 2 α
2sin∧2α-sinαcosα+cos∧2α
=(2sin∧2α-sinαcosα+cos∧2α)/(sin∧2α+cos∧2α)
The denominator is divided by cos Λ 2 α
=(2tan∧2α-tanα+1)/( tan∧2α+1)
=(5-√2)/3.
Let a = {x | x ^ 2-2x-15a ^ 2}. If a ∩ B = & # 8709;, then the value range of positive real number a is?
I don't know what's wrong with that?
A={x|-3
A is greater than or equal to 5
All of the above are true, a ^ 2 ≥ 25 and a > 0
Then a ≥ 5
Known function f (x) = x ^ 2 + ax + B, inequality f (x)
It is known from the title that the two of F (x) = 0 are - 1 and 5
That is, a = - 4; b = 5 · (- 1) = - 5
(1)f(x)=x²-4x-5
(2) F (x) = (X-2) &# 178; - 9 (x ∈ R), and the range of F (x) is [- 9, + ∞]
(3) When x ∈ (- ∞, - 1) ∪ (5, + ∞), | f (x) | = f (x) = x & # 178; - 4x-5; when x ∈ [- 1,5], | f (x) | = - f (x) = - X & # 178; + 4x + 5
1 ° when x ∈ (- ∞, - 1) ∪ (5, + ∞), f (x) and straight line are combined to obtain X & # 178; - 6x - (5 + m) = 0, Δ = 56 + 4m
When Δ ＜ 0, i.e. m ＜ - 14, two functions have no intersection at (- ∞, - 1) ∪ (5, + ∞);
When Δ = 0, i.e. M = - 14, the solution is x = 3, which does not satisfy the assumption that there is no intersection between two functions in (- ∞, - 1) ∪ (5, + ∞);
When Δ＞ 0, i.e. m ＞ - 14, X1 = 3 + √ (14 + m), X2 = 3 - √ (14 + m). Let x2 ＜ - 1, m ＞ - 2; let x1 ＜ 5, m ＞ - 10
When x ∈ (- ∞, - 1) ∪ (5, + ∞): when m ∈ (- ∞, - 10], there is no intersection; when m ∈ (- 10, - 2], there is only intersection; when m ＞ - 2, there are two intersections
2 ° when x ∈ [- 1,5], f (x) and straight line are combined to obtain X & # 178; - 2x + m-5 = 0, Δ = 24-4m
When Δ < 0, i.e. m > 6, the two functions have no intersection at [- 1,5];
When Δ = 0, i.e. M = 6, the solution is x = 1, which satisfies the assumption that there is a unique intersection point between two functions at [- 1,5];
When Δ ＞ 0, i.e. m ＜ 6, X1 = 1 + √ (6-m), X2 = 1 - √ (6-m). Let x2 ≥ - 1, m ≥ 2; let x1 ≤ 5, m ≥ - 10
When x ∈ [- 1,5]: when m ＞ 6 or m ＜ - 10, there is no intersection; when m = 6 or m ∈ [- 10,2), there is a unique intersection; when m ∈ [2,6], there are two intersections
In conclusion, when m ＜ - 10, there is no intersection, when m = - 10, there is a unique intersection, when m ∈ (- 10, - 2] ∪ (6, + ∞), there are two intersections, when m ∈ (- 2,2] or M = 6, there are three intersections, when m ∈ (2,6], there are four intersections
Sn = n ^ 2, let BN = 1 / Anan + 1, TN is the sum of the first n terms of BN, try to prove TN
When Sn = n ^ 2n > = 2, an = SN-S (n-1) = n ^ 2 - (n-1) ^ 2 = 2n-1n = 1, A1 = S1 = 1 also satisfies the above formula BN = 1 / ana (n + 1) = 1 / [(2n-1) (2n + 1)] = (1 / 2) [(2n + 1) - [(2n-1)] / [(2n-1) (2n + 1)] = (1 / 2) [1 / (2n-1) - 1 / (2n + 1)] TN = B1 + B2 + +Bn=(1/2)[(1/1-1/3)+(1/3-1/5)+...
Given that cos (α + β) = 1 / 5, cos (α - β) = 3 / 5, it is proved that sin (2 α + β) / sin α - 2cos (α + β) = sin β / sin α
sin(2α+β)/sinα-2cos(α+β)=sin[(α+β)+α]/sin-2cos(α+β)=[sin(α+β)cosα+cos(α+β)sinα-2cos(α+β)sinα]/sinα=[sin(α+β)cosα-cos(α+β)sinα]/sinα=[sin(α+β-α)]/sinα=sinβ/sinα
Given the set M = {x | x + 2 ≤ 1}, n = {x | x > A-1}, if M ∩ n = & # 8709;, then the value range of real number a is__ sit back and wait
Because M = {x | x + 2 ≤ 1}, X ≤ - 1
And because m ∩ n = &, X is greater than - 1 in x > A-1
So a > 0
If a is greater than o, the first result is that x is less than or equal to - 1. If M intersection n is an empty set, then the second X is greater than - 1, that is, A-1 is greater than - 1.
I'm sorry, I can't see how much m-intersection n equals. If m-intersection n equals infinity, we can get x 2 less than or equal to 1 and X less than or equal to - 1 from the set M. then we can see that the set n is x greater than A-1. We can get A-1 less than - 1 and a less than 0 from the solution. Hope to adopt ～