In the arithmetic sequence {an}, the tolerance D is not equal to. 0, and A1, A2, A5 are in the same ratio sequence. Find the value of a1 + a3 + A9 / A2 + A4 + a10

A1, a1 + D, a1 + 4D into equal proportion sequence
(a1+d)^2=a1(a1+4d)
a1^2+2a1d+d^2=a1^2+4a1d
D ^ 2 = 2a1d D is not equal to 0
d=2a1
a1+a3+a9/a2+a4+a10
=(a1+a1+2d+a1+8d)/(a1+d+a1+3d+a1+9d)
=23a1/29a1
=23/29

Given that the tolerance D of the arithmetic sequence {an} is not equal to 0, and A1, A3 and A9 are equal ratio sequence, then (a1 + A5) / (A2 + A10)?

Know from the title
a1*a9=(a3)^2
Let the tolerance be d
Then A1 * (a1 + 8D) = (a1 + 2D) ^ 2
That is, (A1) ^ 2 + 8a1d = (A1) ^ 2 + 4A1 D + 4D ^ 2
4A1 d = 4D ^ 2
We know that D is not equal to 0
a1=d
So a1 + A5 = 6D
a2+a10=12d
So (a1 + A5) / (A2 + A10) = 1 / 2

It is known that the images of two first-order functions y = 2X-4, y = x + 1 intersect at point a (1) to find the coordinates of point a (2) to find the image of another first-order function and the area of the figure enclosed by X axis

1.2x-4=x+1 x=5 y=6 A(5,6)
2. When y = 0, x = 2 and x = - 1, so the area with X axis = 1 / 2 * 3 * 6 = 9

Given that the image of a function of degree passes through a point (3, - 3) and intersects a line y = 4x-3 on the x-axis, the analytic expression of the function of degree is obtained

We know that the image of a linear function is a straight line, because the straight line y = 4x-3 intersects a point on the X axis, let y = 0, get x = 3 / 4, so the straight line also passes through the point (3 / 4,0),
After two points (3, - 3), (3 / 4,0), there are many ways to solve the linear equation. Here, you can set y = KX + B, and substitute the coordinates of two points respectively, and finally get k = - 4 / 3, B = 1, so the linear equation is y = - (4 / 3) x + 1, that is, the analytic formula of the first-order function is = - (4 / 3) x + 1

We know the first-order function y = KX_ (1) find the intersection of the graph of this linear function with x-axis and y-axis

Substituting the point m (- 2,1) into the function y = kx-3, we get
-2k-3 = 1, k = - 2
So the analytic expression of this function is y = - 2x-3
The intersection coordinates of the image of the function and the x-axis are (- K / B, 0) = (3 / 2,0) and the intersection coordinates of the y-axis are (0, b), that is (0, - 3)

Given that the image of a first-order function passes through two points a (2,4), B (0,2) and intersects with X axis at point C, we can find: (1) the image of a first-order function
(2) the area of △ AOC

Let the analytic expression of a function be y = KX + B
X = 2, y = 4; X = 0, y = 2
We obtain {4 = 2K + B}
2=b
∴﹛k=1,b=2
The analytic formula of a function is y = x + 2
When y = 0, x = - 2
∴C(-2,0)
The area of △ AOC = (?) 189; × 2 × 4 = 4

How many zeros are there in the function y = |log2|x| - 1?

The judgment function y = log2 (x + x ^ 2 - 2) has only one zero point in (1,2)

U = x ^ 2 + X-2 = (x + 2) (x-1) it can be seen that the quadratic function u = (x + 2) (x-1) is an increasing function in the interval (1,2)
And U (1) = 0, u (2) = 4
That is, the quadratic function u = (x + 2) (x-1) has only one X value in the interval (1,2) such that x ^ 2 + X-2 = 1
So the function y = log2 (x + x ^ 2 - 2) has only one zero point in (1,2)

It is known that the definition domain of function f (x) = x2-4x-4 is [T-2, T-1]. For any t ∈ R, the analytic expression of the minimum value g (T) of function f (x) is obtained

F (x) = x2-4x-4 = (X-2) 2-8; if T-1 ≤ 2, i.e. t ≤ 3, f (x) decreases monotonically on [T-2, T-1];; G (T) = f (t-1) = t2-6t + 1; if T-2 < 2 < T-1, i.e. 3 < T < 4, G (T) = f (2) = - 8; if T-2 ≥ 2, i.e. t ≥ 4, f (x) increases monotonically on [T-2, T-1];; G (t

Find the definition domain of function 1. F (x) = 1 / 4x + 7 2. F (x) = √ 1-x + √ x + 3-1

①4X+7 ≠0 X ≠-7/4 [-∞,-7/4) ∪[-7/4,+∞)
②1-X≥0,X+3≥0
X≤1,X≥-3
[-3,1]

In the arithmetic sequence {an}, the tolerance D is not equal to. 0, and A1, A2, A5 are in the same ratio sequence. Find the value of a1 + a3 + A9 / A2 + A4 + a10