80 / 1 / 8 / 1 / 800 * 1 / 800 * 1 / 8000 = how much

80 / 1 / 8 / 1 / 800 * 1 / 800 * 1 / 8000 = how much

80 / 8 * 800 / 8000 = 1

How much is 3.14 times 6 times 80

1507.2

If the median of two positive numbers a and B is 92 and the median of one is 25, and a > b, then the eccentricity of hyperbola x2a2 − y2b2 = 1 is ()
A. 53B. 414C. 54D. 415

According to the meaning of the title, we get a + B = 9ab = 20, the solution is a = 5, B = 4  C2 = A2 + B2 = (a + b) 2-2ab = 41 ｛ C = 41 ｛ e = CA = 415, so we choose D

If the unequal real numbers a, B and C are equal difference sequence, C, a and B are equal ratio sequence, and a + 3B + C = 10, then a=______ ．

Let a = B-D, C = B + D, and let B − D + 3B + B + D = 10 (B − d) 2 = B (B + D) ∵ D ≠ 0, B = 2, d = 6, a = B-D = - 4

a. B, C, D four numbers into an equal ratio sequence, A-1, B-1, C-4, D-13 into an equal difference sequence, find a, B, C, D is how much?

Let the common ratio be q, then these four numbers can be expressed as a, AQ, AQ & # 178;, AQ & # 179;, where AQ ≠ 0, when q = 1, a = b = C = D, then A-1, B-1, C-4, D-13 can't be equal difference, so Q ≠ 1, because A-1, B-1, C-4 can be equal difference, so (A-1) + (AQ & # 178; - 4) = 2 (AQ-1) can get AQ & # 178; - 2aq +

If a, B and C are equal difference sequence, and a, B, C + 3 and a + 1, B and C are equal ratio sequence, then a + B + C =?

There are two equal ratio sequences
b^2=ac+3a=ac+c
So C = 3A
It can be seen from the arithmetic sequence
2b=a+c=4a,
So B = 2A
If B ^ 2 = AC + 3a, a = 3
So B = 6, C = 9
a+b+c=18

If a, 1, C are known to be equal difference sequence and A2, 1, C2 are equal ratio sequence, then loga + C (A2 + C2)=

1,a+c=2,ac=1

(1) A, B, C (a) are known

(1) According to the meaning of the problem, there must be a tolerance d > 0. Let B = A. according to the principle of arithmetic sequence, there are:
Let a = A-D, B = a, C = a + D
If two numbers are exchanged, there are three cases as follows:
1、 Suppose that the two numbers a and C are exchanged. If the three numbers C, B and a are in equal proportion sequence, according to the principle of equal proportion sequence, there should be the following relationship:
a²＝(a－d)(a＋d)
The solution is as follows
d＝0
∵ there is a contradiction between the derived d = 0 and the known condition D > 0
The hypothesis does not hold, and C, B and a are not equal in sequence
2、 Suppose that the two numbers a and B are exchanged. If the three numbers B, a and C are in equal proportion sequence, according to the principle of equal proportion sequence, there should be the following relationship:
(a－d)²＝a(a＋d)
The results are as follows
d²＝3ad
∵ D ≠ 0 (according to the known condition D ＞ 0), both sides of the above equation can be divided by D, and the solution is as follows:
d＝3a
3、 Suppose that the two numbers B and C are exchanged. If the three numbers a, C and B are in equal proportion sequence, according to the principle of equal proportion sequence, there should be the following relationship:
(a＋d)²＝a(a－d)
The results are as follows
d²＝－3ad
∵ D ≠ 0 (according to the known condition D ＞ 0), both sides of the above equation can be divided by D, and the solution is as follows:
d＝－3a
Therefore, according to the above three cases of the exchange of two numbers, only the second and third hypotheses are in line with the meaning of the problem, that is, only when B, a, C or a, C, B are in equal proportion sequence
According to the second and third hypotheses:
d＝±3a
∴（A²＋C²）/B²
＝[(a－d)²＋(a＋d)²]/a²
＝2(a²＋d²)/a²
＝2[a²＋(±3a)²]/a²
＝2(a²＋9a²)/a²
＝20
（2）(1)a(n+1)-an=(n+1+2013)-(n+2013)=1
∴b(n+1)-bn=cn/[a(n+1)-an]=cn=2^n+n
∴bn-b(n-1)=2^(n-1)+n-1
...
b2-b1=2^1+1
Cumulative: bn-b1 = (2 ^ 1 + 2 ^ 2 +.. + 2 ^ (n-1)) + (1 + 2 + 3 +.. + n-1)
=2×(1-2^(n-1))/(1-2)+(1+n-1)(n-1)/2
=2^n-2+n(n-1)/2
∴bn=2^n+n(n-1)/2-1
(2)a(n+1)-an=(n+1)²-8(n+1)-n²+8n=2n-7
∴b(n+1)-bn=n³/(2n-7)
When B (n + 1) - BN > 0, 2n > 7, n ≥ 4
∴b4＜b5＜b6＜..
When B (n + 1) - BN < 0, n ≤ 3
∴b1＞b2＞b3
When∵ n = 3, b4-b3 = 3 & # 179; / (- 1) < 0
∴b4＜b3
B4 is the smallest
∴k=4
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b. M, N, X and y are positive numbers. If a, m, B and X are equal difference series and a, N, B and y are equal proportion series, compare the sizes of M and N, X and y
M is greater than N, but I don't know why x is less than y
Er... The purpose of my question is to understand the process, ha ha

It is known that a, B, m, N, X and y are positive numbers. If a, m, B and X are equal difference sequence and a, N, B and y are equal proportion sequence, compare the size of M and N, X and y
Let the tolerance be D and the common ratio be Q
yes
m=a+d
b=a+2d
x=a+3d
n=aq
b=aq^2
y=aq^3
B = a + 2D = AQ ^ 2 get d = a (Q ^ 2-1) / 2
m-n=a+d-aq=a+a(q^2-1)/2-aq=a(q-1)^2>=0
m>=n
(when q = 1, all six numbers are equal to a)
x-y=a+3d-aq^3=a+(3/2)*a(q^2-1)-aq^3
=(-a/2)(2q^3-3q^2+1)
=(-a/2)(q^2-2q+1)(2q+1)
=(-a/2)(q-1)^2*(2q+1)
a>0,-a/2=0,
2q + 1 > 0 (all items in the equal ratio sequence are positive, and the common ratio must be greater than 0)
So X-Y

It is known that a and B are two unequal positive numbers, and a, x, y and B are in equal difference sequence in turn, and a, m, N and B are in equal proportion sequence in turn. Try to compare x + y and M + n

a. X, y, B in turn into arithmetic sequence
x+y=a+b
a. M, N and B are in equal proportion sequence
mn=ab
m+n>2√(mm)
m+n>2√(ab)
(x+y)-(m+n)
=a+b-(m+n)
0
(x+y)-(m+n)