If M + 1 / M = 3, then the result of square of M + 1 / 2 of M is?

If M + 1 / M = 3, then the result of square of M + 1 / 2 of M is?

(m+1/m)²=3²
m²+1/m²+2=9
m²+1/m²=7

We know that M + 1 / M = 3, find the square of m plus 1 / 2 of M!
Such as the title
emergency

m + 1/m = 3
m^2 + 1/m^2 = (m + 1/m)^2 - 2 = 3^2 - 2 = 7

If M =? M square + 10m + 23 has the minimum value

Square of M + 10m + 23
=(m^2+10m+25)-2
=(m+5)^2-2
Because (M + 5) ^ 2 ≥ 0
So (M + 5) ^ 2-2 ≥ - 2
So when (M + 5) ^ 2, (M's square + 10m + 23) has a minimum value of - 2
m=-5

Let s be a nonempty subset of a real number set R. if there is x + y, XY ∈ s for any x, y ∈ s, then s is called a closed set. Such a closed set s can be____ (give two examples of closed sets)

Such a closed set s can be a positive integer set n *, a rational number set Q 】

Let s be a nonempty subset of real number r. if x, y belong to s, x + y, X-Y, XY belong to s, then s is called a closed set

1. Correct proof: let x, y ∈ s, let x = a + B √ 3, y = C + D √ 3, then x + y = (a + C) + (B + D) √ 3, since a, B, C, D are all integers, then a + C, D + B are also integers, so x + y ∈ SX + y = (A-C) + (B-D) √ 3, since a, B, C, D are all integers, then a-c, D-B are also integers, so X-Y ∈ sxy = AC + 3bd + (AD + BC) √ 3

Let s be a nonempty subset of the complex set C. if there are x + y, X-Y, XY ∈ s for any x, y ∈ s, then s is called a closed set. The following propositions: ① set s = {a + bi | (a, B are integers, I are imaginary units)} is a closed set; ② if s is a closed set, then there must be 0 ∈ s; ③ a closed set must be infinite; ④ if s is a closed set, then any set t satisfying s ⊆ t ⊆ C is also a closed set The truth proposition is______ (write the serial numbers of all true propositions)

The sum of two complex numbers is a complex number, and the difference of two complex numbers is also a complex number, so the set s = {a + bi | (a, B is an integer, I is an imaginary unit)} is a closed set, which is correct. When s is a closed set, because X-Y ∈ s, take x = y, get 0 ∈ s, which is correct. For the set s = {0}, obviously satisfies all conditions, but s is a finite set. ③ wrong, take s = {0}, t = {0, 1}, satisfy s ⊆ t ⊆ C, but because 0-1= -1 does not belong to t, so t is not a closed set

For any real number x, there exists a real number y such that xy = 1, and the non proposition of proposition p is?

For any real number x, there is no real number y such that xy = 1
Non proposition only negates conclusion

(1) Quarter y ^ 2-xy + x ^ 2 (2) - A ^ 4 + 2A ^ 2b-b ^ 2

(1) This is a 188; Y & \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\(b) & 178;

Given the set a = {x, XY, (XY-1) / XY} B = {0, | x | y}, and a = B, find the value of X, y

The definition field XY ≠ 0, because a = B, so (XY-1) / xy = 0 get xy = 1, so a = {x, 1,0} ① if y = 1, substitute xy = 1, the solution get x = 1, then a = {1,1,0} violates the set anisotropy, if | x | = 1, the solution get x = ± 1 (1) x = 1, it is the same as ①, when (2) x = - 1, substitute xy = 1, the solution get y = - 1, then a = {- 1,1,0

Given the set a = {x, XY, X-Y} and the set B = {0, | x | y}, if a = B, find the value of X, y

Because the elements of the set are not repetitive
So there are: B = {0, | x | y}, → x ≠ 0 and Y ≠ 0 and Y ≠ X|
A={x,xy,x-y}→x≠xy≠x-y→x≠0,y≠1,y≠0
Since a = B, there must be: X-Y = 0, that is, x = y, X or Y is negative
So x is not equal to | X|
Xy = | x | = - x → x (y + 1) = 0 → y + 1 = 0 → y = - 1, so x = - 1
The result is: x = y = - 1