It is known that a and B are mutually different positive numbers, a is the mean of the equal difference of a and B, and G is the mean of the equal ratio of a and B, then the size relationship between a and G is () A. A＞GB. A＜GC. A≤GD. A≥G

It is known that a and B are mutually different positive numbers, a is the mean of the equal difference of a and B, and G is the mean of the equal ratio of a and B, then the size relationship between a and G is () A. A＞GB. A＜GC. A≤GD. A≥G

A = a + B2, g = ± AB can be obtained from the meaning of the title, and a ≥ G can be obtained from the basic inequality if and only if a = B takes the equal sign, and a and B are different positive numbers, so a > G, so a is selected

What is the relationship between the size of the mean of the equal difference and the mean of the equal ratio of two positive numbers A.B_____________ .

Proportional median

It is known that a > 0, b > 0, the median of a, B is a, and the median of a, B is g, and G > 0, then the size relationship between AB and Ag

(radical a-radical b) ^ 2 > = 0
A + B-2 radical (AB) > = 0
That is: (a + b) / 2 > = radical (AB)
Multiply both sides by the root sign (AB): get [(a + b) / 2] * [root sign (AB)] > = ab
Notice that the first item on the left is a and the second item is g
So: Ag > = ab

Given that a, B ∈ R +, a is the median of a, B, and G is the median of a, B, then the relation between AB and Ag is ()
A. AB = AGB. Ab ≥ AGC. Ab ≤ AGD

According to the meaning of the title, a = a + B2, g = AB, Ag AB = a + B2, AB AB AB = AB (a + B2 AB) = ab · (a − b) 22 ≥ 0, Ag ≥ ab

Given that AB is a positive number, a is the median of a and B, G is the median of a and B, and a = 2G, find the value of a / b

2A = a + B, G ^ 2 = AB, and a ^ 2 = 4G ^ 2
That is, (a + b) ^ 2 = 16ab
Let a / b = k > 0, a = Bk
b^2(k+1)^2=16b^2*k
(k+1)^2=16k
k^2-14k+1=0
k=7+4√3
k=7-4√3

It is known that ab belongs to R +, a is the mean of equivalency of AB, and G is the mean of equivalency of AB, then the relationship between AB and Ag is? Ab
ab=AG
ab>=AG
ab

ab≤AG
According to the important inequality:
(a + b) / 2 ≥ radical (AB)
That is, a ≥ G
So: Ag ≥ G ^ 2 = ab
Get proof

We know that the median of the equal difference of two numbers a and B is 5. What is the maximum of the median of the equal ratio of a ^ 2 and B ^ 2,

A ^ 2, B ^ 2 are equal in proportion, namely ab
From the basic inequality
√(ab)

Given two numbers 3 and 3x, 3A ^ 2 = 4B ^ 2 is satisfied between a and B
Find a

3-a=a-3x
a=(3x+3)/2
3/b=b/3x
b^2=9x
3a^2=4b^2
48x=(3x+3)^2
3x^2-10x+3=0
(x-3)(3x-1)=0
X = 3
x=1/3
a=(3x+3)/2
a=2

For two given numbers 3 and 3x, if their mean terms a and B satisfy 3A & sup2; = 4B & sup2;, then the positive number B=

A = (3 + 3x) / 2
Proportional median B = √ (3.3x)
Satisfy 3A & sup2; = 4B & sup2;;
Substituting into the solution: x = 3 or 1 / 3, a = 6 or 2, B = 3 √ 3 or √ 3

For two given numbers 3 and 3x, if 3A square equals 4B square, then B equals?

The questions are as follows:
2a=3+3x;
3a^2=4b^2;
b^2=3*3x.
If three equations are solved with three unknowns, B = 3 or 1 / 3 can be obtained,
You can verify that both numbers hold