Set proof: Let the set {a = a | a = 3N + 2, n ∈ Z} and the set B = {B | B = 3K-1, K ∈ Z}, prove that a = B

Set proof: Let the set {a = a | a = 3N + 2, n ∈ Z} and the set B = {B | B = 3K-1, K ∈ Z}, prove that a = B

Let m = k-1, K ∈ Z, then m ∈ Z
We obtain B = 3K-1 = 3 * (M + 1) - 1 = 3M + 2
And because a = 3N + 2, n ∈ Z
So a = B

Let s be a set of at least two elements, and define a binary operation "*" on S (that is, for any a, B ∈ s, for any ordered element pair (a, b), there is a uniquely determined element corresponding to it) with a * (b * a) = b, then for any a, B ∈ s, the following equation does not always hold ()
A．（a*b）*a=a B．[a*（b*a）]*（a*b）=a
C．b*（b*b）=b D．（a*b）*[b*（a*b）]=b
Analytic D says that, if a * B is regarded as a, why can it be like this?
And why are a and B interchangeable,
Why does C only change a into B? I really don't know,

For B, a * (b * a)] = B,. [a * (b * a)] * (a * b) = b * (a * b) = BC, it is obvious that a, B is a positional number, just like x, y does not mean that a, B can be interchanged. The answer to their matter is the same as that in the analysis. If (a * b) = C, then there is C (b * c) = Ba * (b * a) = B

Let the image of function y = f (x) and function y = g (x) be as shown in the figure, then the image of function y = f (x) · g (x) may be ()
A. B. C. D.

From the image of function y = f (x) and function y = g (x), we can see that the image of function y = f (x) is symmetric about y axis, the image of function y = g (x) is symmetric about origin, ∧ function y = f (x) is even, function y = g (x) is odd, ∧ function y = f (x) · g (x) is odd, image is symmetric about origin

Let X and y be positive numbers, and X + y = 1, then the minimum value of a with √ x + √ y ≤ a constant is__________

Radical 2
Because x + y = 1 and x = √ x ^ 2, y = √ y ^ 2
And X + Y > = √ XY, that is, 2 √ XY

At least let me understand)
If f (x) is an odd function on (- 00,0) U (0, + 00) and an increasing function on (0, + 00), f (- 2) = 0, f (- 2) = 0, then the solution set of inequality x * f (x) less than 0 is___________

F (x) is an odd function on (- 00,0) U (0, + 00)
And it is an increasing function on (0, + 00)
It can be concluded that f (x) is an increasing function on (- 00,0) U (0, + 00)
F(-2)=0,
So x * f (x) is equal to 0 when x = - 2
When x is less than - 2, f (x) is less than 0 because it is an increasing function
Negative is positive
So x * f (x) is greater than 0
When x is (- 2,0), f (x) is greater than 0 because it is an increasing function
Negative, positive, negative
So x * f (x) is less than 0
When x is (0,00), f (x) is greater than 0 because it is an increasing function
Zheng Zheng de Zheng
So x * f (x) is greater than 0
So the solution set of inequality x * f (x) less than 0 is: (- 2,0)

As shown in the figure, in the triangular pyramid a-bcd, BC = AC, ad = BD, make be ⊥ CD, e is perpendicular, make ah ⊥ be in H

As shown in the figure, take the midpoint F of AB and connect CF, DF; ∩ BC = AC, ad = BD, ∩ ab ⊥ CF, ab ⊥ DF, CF ∩ DF = f; ∩ ab ⊥ plane CDF, CD ⊂ plane CD; ⊥ CD ⊥ AB, CD ⊥ be, be ∩ AB = B; ∩ CD ⊥ plane Abe, ah ⊂ plane Abe; ∩ CD ⊥ ah, that is, ah ⊥ CD, ah ⊥ be, be ∩ CD = E; ∩ ah ⊥ plane BCD

On line proof of geometry in senior one
In a pyramid P -- ABCD with a rhombic bottom, the angle ABC = 60 degrees, PA = AC = a, Pb = PD = under the root sign, 2A point E on PD and PE: ed = 2:1 on the edge PC, whether there is a point f that makes BF parallel to the plane AEC, please prove it
Let's draw the picture by ourselves. It's OK to draw normally

Take the midpoint m of PE and connect FM, then FM ‖ CE
From EM = 1 / 2PE = ed, e is the midpoint of MD
Connect BM and BD, let BD intersect AC at O, then o is the midpoint of BD
So BM ‖ OE
It is known from (1) and (2) that plane BFM ‖ plane AEC
So BF is parallel to plane AEC

In the cuboid abcd-a1b1c1d1, ab = BC = a, BB1 = B (b > A), connect AC, BC1, cross point B1, make b1e, vertical BC1, cross CC1 to point E, cross BC1 to point Q
(1) AC1 vertical eb1d;
(2) Plane aa1c1c vertical plane eb1d1

(1) Because AB vertical plane bb1cc1
So AB is perpendicular to b1e
And because b1e is perpendicular to BC1 (known)
So b1e is perpendicular to ABC1
So b1e is perpendicular to AC1
(AB = BC Square) because a1c1 is perpendicular to b1d1
So AC1 is perpendicular to d1b1
So AC1 vertical eb1d
(2) B1d1 vertical a1c1 b1d1 vertical Aa1
So b1d1 vertical plane aa1c
So aa1cc1 is perpendicular to eb1d1

Take any point C in the circle, take C as the center of the circle, make the circle tangent to the diameter ab of the circle at point D, two circles intersect at e and F, and prove that EF bisects CD

I think it can be proved that the quadrilateral EDFC is a diamond. As soon as the picture is drawn, it will come out

Go through a point p outside the plane α where △ ABC is, make Po ⊥ α, perpendicular to o, and connect PA, Pb and PC
It is proved that: (1) if PA = Pb = PC, ∠ C = 90 °, then point O is the midpoint of ab
(2) If PA = PA = PC, then point O is the outer center of △ ABC
(3) If PA ⊥ Pb, Pb ⊥ PC, PC ⊥ PA, then point O is the perpendicular of △ ABC

Prove: (2) connect OA, ob, OC, because Po ⊥ α, so Po ⊥ α,. Po ⊥ OA.PO ⊥OB,PO⊥OC;
Because PA = Pb = PC, Po = Po, so OA = ob = OC, so point O is the outer center of △ ABC
(1) In △ ABC, ∠ C = 90 °, OA = ob = OC, then point O is the midpoint of AB side