The arithmetic sequence and the arithmetic sequence have some similar properties (1) If M + n = P + Q, then a (m) + a (n) = a (P) + a (q) (2) If M + n = 2p, then a (m) + a (n) = 2A (P) I don't understand the problem for a long time. It means that this is the formula of the mean of the equal difference. Ask me to list the formula of the mean of the equal ratio? Seek the power

1) If M × n = P × Q, then a (m) × a (n) = a (P) × a (q)
2) If M × n = P & # 178;, then a (m) × a (n) = A & # 178; (P)
An arithmetic sequence is an addition, while an arithmetic sequence is a multiplication
Given that X and y are opposite numbers, and 3x-7y = 10, find the value of x ^ 2005 + 500y
x. Y is opposite to each other, so x + y = 0, that is, x = - Y
And 3x-7y = 10
Then 3x + 7x = 10
x=1,y=-1
x^2005+500y=1^2005+500*(-1)=-499
On analogical reasoning between arithmetic sequence and arithmetic sequence
An is an arithmetic sequence, then BN = (a1 + A2 + a3 +...) An) / N is also equal difference
If cn is an equal ratio sequence and cn is greater than 0, then DN =?
A similar analogy problem has a theorem
An equal ratio is equivalent to an equal difference or BN equal difference is equivalent to 10 ^ (BN) equal ratio
Cn is equal proportion, and incn is equal difference sequence
So indn = (ina1 + ina2 +...) +Inan) / N is an arithmetic sequence
It is reduced to indn = in (A1 * A2 *...) *an)^(1/n)
So DN = (A1 * A2 *...) *An) ^ (1 / N) is an equal ratio sequence
Given that x.y satisfies the equations 2x-7y = 20 and 3x-y = 7, we can find the x power of Y
As soon as possible within 2 days
x=29/19,y=-46/19
You can use the function calculator to calculate the x power of Y
It is known that the coefficients a, B, C of quadratic function f (x) = AX2 + BX + C satisfy the condition a / M + 2 + B / M + 1 + C / M = 1
Where m > 0
It is proved that AF (M / M + 1) < 0
The function f (x) must have zeros in the interval (0,1)
Because f (M / M + 1) = a [M / (M + 1)] ^ 2 + BM / (M + 1) + C
So f (M / M + 1) = m {AM / (M + 1) ^ 2 + B / (M + 1) + C / M}
=m{am/(m+1)^2+1-a/(m+2)}
=m{1-a/[(m+1)^2*(m+2)]}
So when A0, so AF (M / M + 1) < 0
But when a > 0, AF (M / M + 1) is not necessarily < 0
For example, f (x) = 4x ^ 2, satisfying a / (2 + 2) + 0 + 0 = 1
But AF (M / M + 1) = 4 * f (2 / 3) = 64 / 9 > 0
So it should be a / (M + 2) + B / (M + 1) + C / M = 0,
Then f (M / M + 1) = - AM / [(M + 1) ^ 2 * (M + 2)]
So AF (M / M + 1) = - A ^ 2m / [(M + 1) ^ 2 * (M + 2)]
3x-4y-6z=14 2x-8y +2z=15 3x+4y+6z=16
A linear equation of three variables
3x-4y-6z=14 (1)
2x-8y +2z=15 (2)
3x+4y+6z=16 (3)
(1)+(3),
6x=30
X=5
Substituting (1), (2)
15-4y-6z=14 (4)
10-8y +2z=15 (5)
(4)×2-(5),
20-14y=13
z=1/2
X = 5, z = 1 / 2 into (1),
15-4y-3=14
y=-1/2
x=5,y=-1/2,z=1/2
The coefficients a, B and C of quadratic function f (x) = ax ^ 2 + BX + C are not equal to each other
If 1 / A, 1 / B, 1 / C are equal difference sequence, a, B, C are equal ratio sequence, and the maximum value of F (x) in [- 1,0] is - 6
Then a =?
Is there any mistake? I ask you how you know that the axis of symmetry is negative one
First of all, a, B and C must have the same sign, then the axis of symmetry must be on the negative axis, and then we discuss if the axis of symmetry is on the negative axis
-2
Read more books, this is a very basic quadratic function problem, look at quadratic function
-b/2a=-1
1/a-1/b=1/b-1/c
（4ac-b²）/4a=-6
3x + 4Y = 104y-2z = 205x + 2Z = 30, find the values of X, y and Z
3X+4Y=10①
4Y-2Z=20②
5X+2Z=30③
It is noticed that there are - 2Z in ② and + 2Z in ③
② (3) the sum of the two formulas
4y+5x=50④
4y+3x=10①
④-①：2x=40
x=20
Substituting (1)
4y=-50
y=-25/2
Substituting x = 20 into 3
100+2z=30
2z=-70
z=-35
So x = 20
y=-25/2
z=-35
3x+4y=10
4y-2z=20
5x+2z=30
By subtracting Formula 1 from formula 2 from Formula 3, we can get the following results
-2x=-40
So x = 20
Y = - 15
z=-35
3X+4Y=10 (1)
4Y+2Z=30 (2)
5X+2Z=20 (3)
(1) - (2) get
3x-2z=-20 （4）
(3) + (4) get
8x=0
Substituting x = 0 into 4 yields z = 10
Substituting (1) gives y = 5 / 2
To sum up
x=0 y=5/2 z=10
==========
If it is
5x-2z = 20 then... Expand
3X+4Y=10 (1)
4Y+2Z=30 (2)
5X+2Z=20 (3)
(1) - (2) get
3x-2z=-20 （4）
(3) + (4) get
8x=0
Substituting x = 0 into 4 yields z = 10
Substituting (1) gives y = 5 / 2
To sum up
x=0 y=5/2 z=10
==========
If it is
5x-2z = 20 then
3X+4Y=10 (1)
4Y+2Z=30 (2)
5X-2Z=20 (3)
(1) - (2) get
3x-2z=-20 （4）
(3) - (4) get
2x=40
Substituting x = 20 into 4 gives z = 40
Substituting (1) gives y = - 25 / 2
To sum up
X = 20, y = - 25 / 2, z = 40, fold up
3X+4Y=10 ①
4Y-2Z=20 ②
5X+2Z=30 ③
② + 3 is 5x + 4Y = 50 4
① (4) the equations 3x + 4Y = 10
5X+4Y=50
The solution is 2x = 40
X=20
Take x = 20 into ① and ③ to get y = - 12.5
Z = -... Expansion
3X+4Y=10 ①
4Y-2Z=20 ②
5X+2Z=30 ③
② + 3 is 5x + 4Y = 50 4
① (4) the equations 3x + 4Y = 10
5X+4Y=50
The solution is 2x = 40
X=20
Take x = 20 into ① and ③ to get y = - 12.5
Z= -35
So the solution of the equations is x = 20
Y=-12.5
Z=-35
The results have been verified, there is no problem
It is known that the coefficients a, B, C of quadratic function f (x) = ax ^ 2 + BX + C satisfy the condition a / M + 2 + B / M + 1 + C / M = 1
It is known that the coefficients a, B, C of quadratic function f (x) = ax ^ 2 + BX + C satisfy the condition a / (M + 2) + B / (M + 1) + C / M = 1 (where m > 0)
（1）af(m/m+1) ＜0
Because f (M / M + 1) = a [M / (M + 1)] ^ 2 + BM / (M + 1) + C, f (M / M + 1) = m {AM / (M + 1) ^ 2 + B / (M + 1) + C / m} = m {AM / (M + 1) ^ 2 + 1-A / (M + 2)} = m {1-A / [(M + 1) ^ 2 * (M + 2)]} so when A0, so AF (M / M + 1) < 0, but when a > 0, AF (M / M + 1) is not necessarily < 0, for example, f (x) = 4x ^ 2, satisfying a /
1. It is known that 3x = 2Y = 5Z ≠ 0. Find the value of X + 2Y + 3Y / X-Y + Z
Let 3x = 2Y = 5Z = K
Then x = K / 3, y = K / 2, z = K / 5
x+2y+3y/x-y+z=(K/3+K+3K/5)/(K/3-K/2+K/5)=58

RELATED INFORMATIONS