Given m ∈ R, a ＞ B ＞ 1, f (x) = mxx − 1, try to compare the size of F (a) and f (b)

Given m ∈ R, a ＞ B ＞ 1, f (x) = mxx − 1, try to compare the size of F (a) and f (b)

This is where f (a) - f (a) - f (b) (f (a) - f (b) (f (a) - f (b) (f (a) - f (a) - f (a) - f (b) (B-B) (B-B) (B-B − 1 = m (B − a) (A-1) (A-1) (a-mbb − 1 = m (b) (b) (B-B \ (a) - B (b) (B-B − a) (a (a) (a) (a) (a) (a) (a) (a) (a) (a (a) (b) (b) (b) (b) (a) (a) (a) (b) (b) (b) (b) (b) (b) (b) (b) (b) (a) (a) (a) (a) (a) (a) (a) (a) (a) (a) (a) (a) (a) (a) (b) (b) (b) (b) (b) (b) (b) (b) (b) (b) (b(a) > F (b)

It is known that M belongs to R, a > b > 1, and f (x) = MX / (x-1). If we compare the size of F (a) and f (b), can we reduce m directly by quotient comparison method to get a conclusion?

If M = 0, f (a) = f (b) = 0;
If M is not equal to 0, we can compare f (a) and f (b) by quotient comparison
When f (a) / F (b) = ma / (A-1) / [MB / (B-1)] = a (B-1) / [b (A-1)] = (AB-A) / (ab-b) 0, 0

What is the number of different sets m satisfying the condition that {1,2,5} is the proper subset of M, and M is a = {1,4,8, x, y, X-Y} the proper subset of M?
I don't understand the answer. I don't know which {1,4,8,2,5,7} came from

The {1,4,8,2,5,7} means that the condition {1,2,5} is a proper subset of M, and the condition m is a = {1,4,8, x, y, X-Y} proper subset, so it is right, where x = 7, y = 5, X-Y = 2

Set a = {x | - 2 ≤ x ≤ 5} set B = {x | - M + 1 ≤ x ≤ 2m-1} and B is a subset of A?

If B is an empty set
Then M + 1 > 2m-1
m

Let a = | x | - 3 ≤ x ≤ 4 |, | x | - 2m-1 ≤ x ≤ m + 1 |, when B is a subset of a, the value range of real number M

Let a = | x | - 3 ≤ x ≤ 4 |, B = | x | - 2m-1 ≤ x ≤ m + 1|
When B is a subset of a
1. When B is an empty set, the solution 2m-1 > m + 1 is m > 2
2. If B is not an empty set, 2m-1 ≤ m + 1, m ≤ 2
And 2m-1 ≥ - 3M + 1 ≤ 4
The solution is m ≥ - 1 m ≤ 3
So - 1 ≤ m ≤ 2 (binding condition)
In conclusion: m ≥ - 1

If a = {x | x > 2 or X

If B is a subset of a, then there are:
a> = 2 or a + 1 = 2 or a

If a = {x | x ^ 2-5x + 6 = 0}, B = {x | x ^ 2-4ax + 3A ^ 2}

The solution of a is x = 2 or x = 3
The solution of B is X1 = a, X2 = 3A
x1

Let a = {x | x ^ 2-5x-6 ≥ 0}, B = {x | 2a-1}

Because set a = {x | x ^ 2-5x-6 ≥ 0},
So set a = {x | x ≤ - 1 or X ≥ 6},
Because a ∩ B = empty set, 2a-1 ≥ - 1, 3a-2 ≤ 6, 2a-1 < 3a-2
Solve three inequalities: 1 < a ≤ 8 / 3

Known set a = {xlax ^ 2 + 2x + 1 = 0}
1: If a = an empty set, find the value of A. 2: if there is only one element in a, find the value of A. 3: if there is at most one element in a, find the value of A,

1. Set a = {xlax ^ 2 + 2x + 1 = 0} = empty set
ax^2+2x+1=0
4-4a1
2. Set a = {xlax ^ 2 + 2x + 1 = 0} = an element
ax^2+2x+1=0
4-4a=0
a=1
3. Set a = {xlax ^ 2 + 2x + 1 = 0} has at most one element
ax^2+2x+1=0
4-4a=1

Given that real numbers x and y satisfy x ^ 2 + y ^ 2 + 2x-2 √ 3Y = 0, then the maximum value of (y + √ 3) / (x-1) is obtained

Let (y + √ 3) / (x-1) = k, then y = kx-k - √ 3
(1 + K ^ 2) x ^ 2-2 (k ^ 2 + 2 √ 3K-1) x + K ^ 2 + 4 √ 3K + 9 = 0
k