Let a (x1, Y1), B (X2, Y2) be functions f (x) = (1 / 2) + log2 (x / 1-x), and OM = (1 / 2) (OA + OB) The abscissa of point m is 1 / 2, and Sn = f (1 / N) + F (2 / N) + +F (n-1 / N), where n ∈ n * and N ≥ 2, (1) Calculate the ordinate value of point M; (2) Find S2, S3, S4 and Sn; (3) Given an = 1 / (Sn + 1) (Sn + 1 + 1), where n ∈ n *, and TN is the sum of the first n terms of the sequence {an}, if TN ≤ λ (Sn + 1 + 1) holds for all n ∈ n *, try to find the minimum positive integer value of λ

Let a (x1, Y1), B (X2, Y2) be functions f (x) = (1 / 2) + log2 (x / 1-x), and OM = (1 / 2) (OA + OB) The abscissa of point m is 1 / 2, and Sn = f (1 / N) + F (2 / N) + +F (n-1 / N), where n ∈ n * and N ≥ 2, (1) Calculate the ordinate value of point M; (2) Find S2, S3, S4 and Sn; (3) Given an = 1 / (Sn + 1) (Sn + 1 + 1), where n ∈ n *, and TN is the sum of the first n terms of the sequence {an}, if TN ≤ λ (Sn + 1 + 1) holds for all n ∈ n *, try to find the minimum positive integer value of λ

1. F (x) = (1 / 2) + log2 (x / (1-x)) om = (OA + OB) / 2 = (x1 + X2, Y1 + Y2) / 2 = (x1 + x2,1 + log2 {x1x2 / [(1-x2) (1-x2)]}) / 2 = (X1 + x2,1 + log2 [x1x2 / (1-x1-x2 + x1x2)]) / 2 if the abscissa of the known point m is 1 / 2, then X1 + x2 = 1, and OM = (1,0) / 2 = (1 / 2,0) is obtained

It is known that: as shown in the figure, four rays OA, ob, OC and od are introduced from point O, if OA ⊥ OC, ob ⊥ OD. (1) if ∠ BOC = 35 °, calculate the size of ∠ AOB and ∠ cod; (2) if ∠ BOC = 50 °, calculate the size of ∠ AOB and ∠ cod; (3) what is the relationship between ∠ AOB and ∠ cod?

(1) ∫ OA ⊥ OC, ∫ AOB + ∠ BOC = 90 °, ∫ BOC = 35 °, ∫ AOB + 35 ° = 90 °, ∫ AOB = 55 ° can be obtained by the same principle: ∫ cod = 55 ° (2) ∫ OA ⊥ OC, ∫ AOB + ∠ BOC = 90 °, ∫ BOC = 50 °, ∫ AOB + 50 ° = 90 °, ∫ AOB = 40 ° can be obtained by the same principle: ∫ cod = 40 °; (3) from the calculation of (1) and (2), we can know: ∫ AOB = cod

Known: as shown in the figure, starting from the point O, five rays OA, ob, OC, OD and OE are introduced, with ∠ AOB = 90 °
(1) If ∠ BOD is equal to 3 times of the complementary angle of ∠ AOD, calculate the degree of ∠ AOD;
(2) If OE bisects ∠ AOD, and ∠ AOC: ∠ BOC = 1:2, the degree of ∠ EOD is required, but there is still one condition to be left. Now we give the following conditions: ① ∠ AOC = 30 °, ② ∠ BOC = 60 ° and ③ ∠ cod = 166 °. Can you choose one of these three conditions (only one can be selected), and add the previous conditions to find out the degree of ∠ COE?
It's not easy for me to draw a picture, but when I look clockwise, the order of rays is ob, OC, OA, OE, OD. It seems that ∠ boa is a right angle. That's all I can give you. Have a knowledge of grade one,

Solution 1) ∠ BOD + ∠ AOD = 360 °∠ BOD is equal to 3 times of the complementary angle of ∠ AOD,
Then ∠ BOD = 3 (180 ° - AOD)
So 3 (180 ° - AOD) + AOD = 360 ° and the solution is ∠ AOD = 90 °
2) Select ③ ∠ cod = 166 °
From ∠ AOC: ∠ BOC = 1:2, ∠ AOB = 90 °, we can get: ∠ AOC = 30, ∠ BOC = 60
Then ∠ BOD = ∠ cod - ∠ BOC = 106,