If the solution set X ≥ 3 of the inequality system {4x + 1 ≥ x + 10 x > m + 1 about X, then the value range of M is ()?

If the solution set X ≥ 3 of the inequality system {4x + 1 ≥ x + 10 x > m + 1 about X, then the value range of M is ()?

Fill in:
m≤2
4x+1≥x+10
3x≥9
x≥3
x＞m+1
If the solution set X ≥ 3, take the larger one according to the large one
m+1≤3
m≤2
If x + 8 is less than 4x-1, X is less than m, and the solution set is x > 3, the value range of M is obtained
Is x > m
x+89
X>3
X>m
The solution set is x > 3
Then 3 ≥ M
So m ≤ 3
There is a problem with the title
Set a = {x | (x ^ 2) - 3x + 2 = 0}, B = {x | (x ^ 2) - ax + A-1 = 0}, whether there is a real number a such that a ∪ B = R, why not?
It certainly doesn't exist
Set a is an equation with at most two elements
Set B is an equation with at most two elements
A ∪ B then there are only four elements at most
Impossible = R
Given the set a = {x 2-3x + 2 = 0}, B = {x 2-ax + A-1 = 0}, ask: is there a so that B is the true subset of a, if there is, find all the values of a; if not, explain the reason
∵ set a = {x 2-3x + 2 = 0} = {1, 2}, solve the equation x2-ax + A-1 = 0 to get: x = 1, x = A-1, if B is the proper subset of a, then A-1 = 1, and a = 2
It is known that the function f (x) is an odd function with the domain R, and its image is symmetric with respect to the straight line x = 1. (1) find the value of F (0). (2) prove that the function f (x) is a periodic function
(1) Because the function f (x) is an odd function with the domain R, f (- x) = - f (x). When x = 0, f (- 0) = - f (0), so f (0) = 0. (2) because the function is symmetric with respect to x = 1, f (1 + x) = f (1-x), that is, f (1 + x) = f (1-x) = - f (x-1), so f (x + 2) = - f (x)
Given that the point (1,2) is a point on the image of the function f (x) = ax (a > 0, a ≠ 1), the sum of the first n terms of the sequence {an} is Sn = f (n) - 1
Given that the point (1,2) is a point on the image of the function f (x) = ax (a > 0, a ≠ 1), the sum of the first n terms of the sequence {an} is Sn = f (n) - 1
(1) Find the general term formula of sequence {an};
(2) If BN = logaan + 1, find the first n terms and TN of the sequence {an &; BN}
How do you get the result that an = sn-sn-1 = 2 ^ n-2 ^ (n-1) = 2 ^ (n-1)
Given that the point (1,2) is a point on the graph of the function f (x) = ax (a ＞ 0, a ≠ 1), we can get a * 1 = 2, that is, the sum of the first n terms of a = 2 sequence {an} is Sn = f (n) - 1, we can get Sn = 2 * n-1, then s (n-1) = 2 * (n-1) - 1An = SN-S (n-1) = (2 * n-1) - [2 * (n-1) - 1] = 2A1 = S1 = f (1) - 1 = 2 * 1-1 = 1
need
Let the domain of F (x) be r, and the graph of F (x) be symmetric with respect to the line x = A and x = B (b > A). It is proved that f (x) is a periodic function with period 2 (B-A)
Given that the function f (x) = x2 + 2x, the sum of the first n terms of the sequence {an} is Sn, for all positive integers n, the point PN (n, Sn) is on the image of function f (x), and the slope of the tangent passing through the point PN (n, Sn) is kn. (I) find the general formula of the sequence {an}; (II) if BN = 2Kn · an, find the sum of the first n terms of the sequence {BN} as TN; (III) let q = {x | x = kn, n ∈ n *}, r = {x | x = 2An, n ∈ n *}, equal difference Any term cn ∈ Q ∩ r of sequence {CN}, where C1 is the least decimal in Q ∩ R, 110 & lt; C10 & lt; 115, to find the general formula of {CN}
(I) point PN (n, Sn) is all in the image of the function f (x) = x2 + 2x, and \\\\\\\\\\ \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\max = 2Kn •an，∴bn＝22n+2•(2n+1)＝4(2n+1)•4n．∴Tn＝4×3×4+4×5×42+4×7×43+… +4（2n+1）•4n… ① It can be concluded that 4tn = 4 × 3 × 42 + 4 × 5 × 43 + 4 × 7 × 44 + +4（2n+1）•4n+1… ② The results show that: (1) T 4 + 4 · + 4 2 · + (2) T 4 + 4 2 · + (3) T 4 + 4 2 · + (3) T 4 + 4 · + (4) T 4 + 4 4 · + 4 · + 4 · + 4 · + 4 · + 4 · +(2n + 1) [(2n + 1) - 4N + 1 (4N + 1) - 4N + 1] = 4 [3 × 4 + 2-42 (1-4n-4n-1) 1 [4 - (2n + 1) - 4N - (2n + 1) [(2n + 1) - 4N (2n + 1) - 4N + 1 (4N + 1) - 4N + 1] = 4 {[3 [3 × 4 + 2, 42 (1-4n-4n-4n-4n-4n-4-42 (1-4n-4n-4n-4n-1) 4 (1-4-42 (1-4n-4n-4n-4n-4n-4n-4n-4-4n-4n-4-4n-42 (1) - 4-4-4n-4n-4n-4n-4n-4n-4-4-4-4-4-4n-4-4-4-4-4-for example, m ∈ n *, ({CN} tolerance is a multiple of 4 & nbsp;) and ∵ 110 & lt; C10 & lt; 115 ∵ 110 & lt; 4m + 6 & lt; 115m ∈ n *. The results show that M = 27, the tolerance of {CN} is 12, CN = 12n-6
It is known that the function f (x) is an odd function with the domain R, and its image is symmetric with respect to the straight line x = 1. (1) find the value of F (0). (2) prove that the function f (x) is a periodic function
(1) Because the function f (x) is an odd function with the domain R, f (- x) = - f (x), when x = 0, f (- 0) = - f (0), so f (0) = 0. (2) because the function is symmetric with respect to x = 1, f (1 + x) = f (1-x), that is, f (1 + x) = f (1-x) = - f (x-1), so f (x + 2) = - f (x), that is, f (x + 4) = f (x). Therefore, the function is a periodic function with a period of 4
Given the quadratic function f (x) = x ^ 2, the sum of the first n terms of the sequence {an} is Sn, and the points (n, Sn) (n ∈ n *) are all on the image of function f (x)
Given the quadratic function f (x) = x ^ 2, the sum of the first n terms of the sequence {an} is Sn, and the points (n, Sn) (n ∈ n *) are all on the image of function f (x)
(1) Find the general term formula of sequence {an}
(2) Let the sequence {BN}, BN = an × 2 ^ n, tn be the sum of the first n terms of the sequence {BN}, and find TN
an=Sn-Sn-1=2n-1
Why is sn-sn-1 equal to 2N-1?
How to write the following steps?
Note: a (n-1) represents the item (n-1) in {{} in the (n-1); 58969; s (n-1) represents the preceding (n-1) items and the (n-1) items of {{{{}} and the (n-1) represents the (n-1) power of the (n-1) power of 2 (n-1) of 2; (n-1) point (n, Sn) on the function f (x) image, f (x) on the function f (x) image, f (x) on the function f (x) image, f (x) image, f (x) = f (x (x (x) = x ^ 2