If the complex Z satisfies | Z + 4 + 3I | = 3, then the module of complex Z should satisfy the inequality? A | Z|

If the complex Z satisfies | Z + 4 + 3I | = 3, then the module of complex Z should satisfy the inequality? A | Z|

A picture is the clearest
|Z + 4 + 3I | = 3, the trajectory of Z is a circle with a (- 4, - 3I) as the center and 3 as the radius
OA=5
It can be seen from the figure that 5-3 ≤| Z | ≤ 5 + 3, 2 ≤| Z | ≤ 8
Choose C
|Z + 4 + 3I | = 3, which means that the distance from point Z to point P (- 4, - 3) in the complex plane is 3, that is, P (- 4, - 3) is a circle with center and radius of 3
And op = 5
|Z | is the distance from Z to the origin, so there is | op | - R=
What does 2x & # 178; mean? Is it (2x) &# 178; or 2 times X & # 178; if x = 3, then 2x & # 178; equals 18 or 36
The latter, without brackets, is 2 * x square, equal to 18
It is known that the general term formula of sequence {an} is an = 2 ^ (2n-1) and BN = Nan, and the first n terms and Sn of sequence {BN} are obtained
This is a typical subtraction problem!
First write s (n) = B (1) +;;;;;;;;;; + B (n) = 2 + 2 * 2 ^ 3 +. + n * 2 ^ (2n-1)
Then write the square of 2 times s (n) = 1 * 2 ^ 3 +. + (n-1) * 2 ^ (2n-1) + n * 2 ^ (2n + 1),
By subtracting the two, we can get 3S (n) = 2 + 2 ^ 3 + 2 ^ 5 +. + 2 ^ (2n-1) - N2 ^ (2n + 1). Except for the last term, the rest is an equal ratio sequence, and the common ratio is 2 ^ 2. Just write the first n terms and subtract the last term!
You can only figure out the basic way of thinking when you write it!
So the sum of the first n terms of B (n) is s (n) = 2.44.46.42n · 4 ^ n (1) 4S (n) = 2.44.42 (n-1) · 4 ^ n 2n · 4 ^ n (n 1)
What a classic question.
It's the sum of the first n terms of an plus N * n
When x is greater than 0, the minimum value of X + 1 / X is 0
Using the formula x ^ 2 + y ^ 2 ≥ 2XY
Because x > 0
X + 1 / X ≥ 2 radical (x * 1 / x)
x+1/x≥2
So the minimum is 2
X + 1 / X
=x+1/x
≥2√x*(1/x)=2
X + 1 / X ≥ 2
When x = 1 / x, x + 1 / X is the smallest.
Here x = 1, x + 1 / x = 2
Minimum 2
The answer is 2
Why don't you tell me how you got it
It's simple
Check the basic inequality on the Internet and you'll get it
Hello:
The basic problem can be solved by using the inequality
X + 1 / x > = 2 √ x × 1 / x = 2
But it depends on the range of X
When x = 1 / x, the original inequality holds, then x = ± 1
So x = 1
If x > 0, then x > 0.
The minimum value is 2
If you don't understand, you can ask.
I wish you progress in your study!
Let u = R, a = {x | 2x-10 ≥ 0}, B = {x ︱ X & # 178; - 5x ≤ 0, and X is not equal to 5}. Find a ∪ B, CUA, Cu (a ∩ b)
Let u = R, a = {x | 2x-10 ≥ 0}, B = {x ︱ X & # 178; - 5x ≤ 0, and X is not equal to 5}. Find a ∪ B, CUA, Cu (a ∩ b)
A∪B={x|x≥0} ,CuA= {x|x
It is known that the sum of the first n terms of {an} is Sn, if A1 = 1, Sn = nan-n (n-1), let BN = 1 / an * an + 1, and the sum of the first n terms of the sequence is TN
1. Prove that the sequence {an} is an arithmetic sequence, and write the expression of {an} about n;
2. If the inequality λ TN ＜ (n + 8) / 5 (n is a constant) holds for any positive integer n, find the value range of λ;
3. Whether there are positive integers m, n (1 ＜ m ＜ n) such that T1, TM, TN are in equal proportion sequence? If so, find out all the values of M, N; if not, explain the reason
1.Sn=nan-n(n-1).1
So s (n + 1) = (n + 1) a (n + 1) - n (n + 1). 2
Formula 2 minus Formula 1: a (n + 1) = (n + 1) a (n + 1) - nan-2n
So Na (n + 1) - Nan = 2n, that is a (n + 1) - an = 2
So an = 2N-1
2.bn=1/[(2n-1)(2n+1)]
Tn=1/(1*3)+1/(3*5)+.+1/[(2n-1)(2n+1)]
=1/2(1-1/3+1/3-1/5+.+1/(2n-1)-1/(2n+1))
=(1/2)*2n/(2n+1)
=n/(2n+1)
That is, λ n / (2n + 1)
Find: when x is greater than or equal to 0, find the minimum value of e ^ x + 1 / x + 4x
e^x+1/x+4x≥e^x+4
It holds when 1 / x = 4x
4x^2=1
X ^ 2 = 1 / 2 (- 1 / 2 rounding off)
And e ^ x increases monotonically
The minimum value is
4+√e
Given that x is not equal to 0, and M = (X & # 178; + 2x + 1) (x-2x + 1), n = (X & # 178; + X + 1) (X & # 178; - x + 1), try to explore the size relationship between M and n!
In your question, m should be copied wrong, which should be (X & # 178; + 2x + 1) (X & # 178; - 2x + 1) M-N = (X & # 178; + 2x + 1) (X & # 178; - 2x + 1) - (X & # 178; + X + 1) (X & # 178; - x + 1) = [(X & # 178; + X + 1) + x] (X & # 178; - 2x + 1) - (X & # 178; + X + 1) [(X & # 178; - 2x + 1) + x]
What's wrong with the title? If there is m = (x ^ 2 + 1) ^ 2-4x ^ 2 n = (x ^ 2 + 1) ^ 2-x ^ 2, it is obvious that M is less than N. question: (X & # 178; + 2x + 1) (X & # 178; - 2x + 1) can be equal to (x + 1) &# 178; (x-1) &# 178
The sum of the first n terms of the sequence an is Sn and an + Sn = - 2N-1. If the sequence BN satisfies B1 = 1 and B (n + 1) = BN + Nan, the general formula of BN can be obtained
an+Sn= -2n-1 a1=-3/2 a1+1=-1/2
an-1+sn-1=-2n+1
2an-an-1=-2
2an+2=an-1+1-1
an+1=1/2(an-1+1)-1/2
1/2(an-1+1)=1/4(an-2+1)-1/4
.
1/2^(n-2)(a2+1)=1/2^(n-1)(a1+1)-1/2^(n-1)
an+1=1/2^(n-1)(-1/2)-(1-1/2^(n-1))=-1/2^n-1+2/2^n
an=-2+1/2^n
b(n+1)=bn+nan=bn+n/2^n-2n
bn=b(n-1)+(n-1)/2^(n-1)-2(n-1)
b2=b1+1/2-2
2+...+2(n-1)=2(n-1+1)(n-1)/2=n(n-1)
s=1/2+2/2^2+3/2^3+..+(n-1)/2^(n-1)
s/2=1/2^2+2/2^2+3/2^3+...+(n-2)/2^(n-1)+(n-1)/2^n
s/2=1/2+1/2^2+1/2^3+...+1/2^(n-1)-(n-1)/2^n=1-1/2^(n-1))-(n-1)/2^n=1-(n+1)/2^n
s=2-(n+1)/2^(n-1)
bn=b1-n(n-1)+2-(n+1)/2^(n-1)=3-n(n-1)-(n+1)/2^(n-1)
Given that x is greater than 0, y is greater than 0 and 4 / x + 9 / y is equal to 2, find the minimum value of X + y
4/x+9/y=2
(4y+9x)/xy=2
4y+9x=2xy
2xy>=2√(4y*9x)=12√xy
2xy-12√xy>=0
2√xy(√xy-6)>=0
√ XY > = 6 or √ xy = 2 √ xy = 2 * 6 = 12