In the known arithmetic sequence {an}, an ≠ 0, tolerance D ≠ 0, (1) prove the equation anx ^ 2 + 2A (n + 1) x + 2A (n + 2), (2) Let the other root of the equation in (1) be {BN}, and prove that {1 / (BN + 1)} is an arithmetic sequence

In the known arithmetic sequence {an}, an ≠ 0, tolerance D ≠ 0, (1) prove the equation anx ^ 2 + 2A (n + 1) x + 2A (n + 2), (2) Let the other root of the equation in (1) be {BN}, and prove that {1 / (BN + 1)} is an arithmetic sequence

It is proved that (1) ∵ {an} is a sequence of arithmetic numbers, 2A (K + 1) = AK + a (K + 2), so the equation akx ^ 2 + 2ak + 1x + AK + 2 = 0 can be changed to [akx + a (K + 2)] (x + 1) = 0, when k takes different natural numbers, the original equation has a common root - 1 (2) and the other root is BN = XK = - A (K + 2) / AK = (AK + 2D) / AK = - 1-2D / AK ∵ 1 / (...)

In the arithmetic sequence {an}, the tolerance d > 0, and A2, A5 are the two (1) general terms of {an} in the equation x ^ 2-6x + 8 = 0
(2) If A2 and A5 are the first two terms of the equal ratio sequence {BN}, find the first six terms of the sequence {BN} and S6

(1) Solving the equation, A2 = 2, A5 = 4, so the tolerance d = (a5-a2) / (5-2) = 2 / 3, so the general term an = A2 + (n-2) d = (2n + 2) / 3. (2). From (1), B1 = 2, B2 = 4, so the common ratio q = B2 / B1 = 2, so BN = 2 ^ n, then S6 = 2 * (1-2 ^ 6) / (1-2) = 2 ^ 7-2 = 126

It is known that the sequence {an} is an arithmetic sequence, the tolerance D ≠ 0, and A1 and A2 are two of the equation x with respect to x, then what is an
Need specific problem-solving process, thank you

From Veda's theorem, we get a1 + A2 = A3 and A1 * A2 = A4, that is, A1 = D and A1 * (a1 + D) = a1 + 3D, and the solution is A1 = 2, d = 2
So an = 2n

The monotone decreasing interval of function y = arccos (x ^ 2-2x) is

Because y = arccos (x) is defined as [- 1,1] and monotonically decreasing in the domain,
Therefore, to find the monotone decreasing interval of function y = arccos (x ^ 2-2x) is to find the increasing interval of function f (x) = x ^ 2-2x
It is easy to know that the increasing interval of F (x) = x ^ 2-2x is x > = 1, in addition to satisfying - 1

If the function f (x) = a'2x-4 '(a > 0, a is not equal to 1) satisfies f (x) = 1 / 9, then what is the monotone decreasing interval of F (x)?
Help The one above is the | 2X-4 | power of F (x) = a

2 to prove infinite

We know the cube of function f (x) = 2x - the cube of ax (a? R and a is not equal to 0, R) (1) find the decreasing interval of function f (x). (2) if not equal
Given the cube (a? R) of function f (x) = 2x - ax and a is not equal to 0, R) (1) find the decreasing interval of function f (x). (2) if the square of inequality x is - 5x + 4

(1),f'(x)=3x(2x-a)
a>0
0

Given that the function f (x) = ax + 1 x + 2 is an increasing function in the interval (- 2, + ∞), then the value range of real number a ()
A. a＞12B. a≤−12C. a≤12D. a≥-12

F ′ (x) = a (x + 2) − (AX + 1) (x + 2) 2 = 2A − 1 (x + 2) 2, because f (x) is an increasing function on (- 2, + ∞), so f ′ (x) ≥ 0 is constant, that is, 2a-1 ≥ 0, the solution is a ≥ 12, and when a = 12, f (x) = 12 is not monotone, so the value range of real number a is a > 12, so select a

It is known that the function f (x) = ax + 1 / x + 2 is an increasing function in the interval (- 2, + ∞). To find the value range of a, is the interval given in this question as long as it is
The answer is that a is greater than half. That is to say, as long as the coefficient of the inverse proportional function is negative, then it must be reduced

f（x）=(ax+1)/(x+2)
=[a(x+2)+1-2a]/(x+2)
=a+(1-2a)/(x+2)
If f (x) is an increasing function in the interval (- 2, + ∞)
Then for any - 2

If f (x) = (AX + 1) / (x + 2) is an increasing function in the interval (- 2, + ∞), then the value range of a is?

Method 1: F (x) = (AX + 1) / (x + 2) = [a (x + 2) - 2A + 1] / (x + 2) = a + (1-2a) / (x + 2). Let y = 1 / (x + 2), and this function is a decreasing function on X ∈ (- 2, + ∞). Now, if y = (1-2a) / (x + 2), and is an increasing function on X ∈ (- 2, + ∞), then it must satisfy (1-2a) 1 / 2

If the monotone interval of the absolute value of the function f (x) = 2x + A is greater than or equal to 3 to positive infinity, then a =?
I'm stupid,
It's a monotone increasing interval

First of all, we show that the image of such a function f (x) = | ax + B | is V-shaped or inverted V-shaped
F (x) = | 2x + a | is a V-shaped image
Let f (x) = 0 and get x = - A / 2
In terms of image, you should draw your own picture. Just draw a V-shape, which should be symmetrical
It can be seen from the figure that the function f (x) is a decreasing function in the interval (- ∞, - A / 2) and an increasing function in the interval [- A / 2, + ∞)
As long as [3, + ∞) is a subset of [- A / 2, + ∞), that is - A / 2 ≤ 3, the solution is a ≥ - 6