Expand the power series (x) / (X-2) = 4x-1

Expand the power series (x) / (X-2) = 4x-1

Use the known expansion formula to calculate, as shown in the figure. The economic mathematics team will help you solve it. Please evaluate it in time
Let the solution set of inequality (2a-b) x + 3a-4b < 0 be x < 49, and find the solution of inequality (a-4b) x + 2a-3b > 0
We can get the solution of {2A} {2A} {4B = 584a-b} by substituting {2A}} {2A}} {2A}} {4B = 584a-0
The equation of the line passing through the focus of the parabola y2 = 2x and parallel to the line 3x-2y + 5 = 0 is ()
A. 6x-4y-3=0B. 3x-2y-3=0C. 2x+3y-2=0D. 2x+3y-1=0
According to the meaning of the question, let the equation of the parallel straight line be 3x-2y + C = 0, and the straight line pass through the focus (12, 0) of the parabola y2 = 2x, and substitute it to get C = - 32, so the equation of the straight line is 6x-4y-3 = 0
Expand f (x) = arctan [(1 + x) / (1-x)] into a power series of X
F '(x) = 1 / (1 + x ^ 2) = ∑ (n = 0 to ∞) (- 1) ^ n · x ^ (2n) x ∈ (- 1,1)
As the sum of equal ratio sequence, why multiply by (- 1) ^ n, I don't understand
This is because the common ratio of the equal ratio sequence is different
1/(1-x) = 1 + x + x^2 + ...+ x^n + ...
1/(1+x) = 1 - x + x^2 + ...+ (-1)^n * x^n
Just replace the second formula x with x ^ 2
Do not multiply by (- 1) ^ n,
Σ (n = 0 to ∞) x ^ (2n) = 1 / (1-x ^ 2) x ∈ (- 1,1),
Note that the series converges to 1 / (1 - x ^ 2) instead of 1 / (1 + x ^ 2), can you see clearly, a positive and a negative?: Thank you, understand
a. B is a rational number, (2a-b) x + 3a-4b
I'm a student of class 3, grade 2, Tongmei north school, Datong, Shanxi
Calculation problem: find the vertex, focus coordinate and quasilinear equation of the square of the parabola 4x + 3Y = 0
Attached problem solving process
x^2=-(3/4)y
x^2=-2py
Then p = 3 / 8
The fixed point is the origin
Opening down
So the focus is on the y-axis
The ordinate is equal to 0-P / 2
So focus (0, - 3 / 16)
So the guide line is y = 3 / 16
The function f (x) = 1x2 + 4x + 3 is expanded into a power series of (x-1)
Because f (x) = 1x2 + 4x + 3 = 1 (x + 1) (x + 3) = 12 (1 + x) − 12 (3 + x) = 14 (1 + X − 12) − 18 (1 + X − 14), and because 11 + x = ∞ n = 0 (− 1) NxN, - 1 ＜ 1, so in the − 1 < x ＜ 3, 14 (1 + X-12) = 14 (1 + X-12) = 14 (1 + X-12) = 14 (1 + X-12) = 14 (1 + X-12) = 14 (14 (1 + X − 12) = 14 (1 + X-12) = 14 (14 (1 + x-14) = 18 (1 + x-14) = 18 (1) nx4n (x (- 1) n (x-1) n) n (x ＜ 1) n4n (x ＜ 1) n, and then notice (- 1, 3) (- 3, 3, 5) = (- 3, 5, 5) = (- 3, 5 (- 3, 5, 5, 5 (- 1, 5, 5, 5, 5-3,5) (- 3, 5) = (- 3, 5 (- 1, 5) = (- 1,5) = (- 1, N, (- 1 < x < 3)
The solution of inequality ax + 3 > 3A about X
ax+3>3a
ax>3a-3
(1)
a> 0, x > (3a-3) / A, x > 3 - 3 / A
(2)
When a = 0, the inequality becomes 0 > - 3, the inequality holds and the solution set is all real number set R
(3)
A
ax+3>3a
ax>3a-3
Obviously, when a = 0, 0 > 0-3 = - 3 holds
When a > 0, the solution is: x > (3a-3) / A
When A0
The solution is x > 3-3 / A
When a3a-3
x>3(a-1)/a
When a3a-3
X3a
ax>3（a-1)
x>3（a-1)/a (a>0)
X
Given that the circle P: x2 + y2-2y-3 = 0, the parabola C takes the center P as the focus and the coordinate origin as the vertex. (1) find the equation of the parabola C; (2) let the intersection of the circle P and the parabola C in the first quadrant be a, the intersection of the tangent of the parabola C through a and the Y axis be q, and the sum of the distances from the moving point m to P and Q is equal to 6, then find the trajectory equation of M
(1) The circle x2 + y2-2y-3 = 0 is transformed into the standard equation: x2 + (Y-1) 2 = 4 ∥ the center of the circle P (0, 1) ∥ Let C: x2 = 2PY (2 points), ∵ parabola C focuses on the center P, P2 = 1 The equation of parabola is x2 = 4Y (2) y = 1 can be obtained from the equations x2 + Y2 − 2Y − 3 = 0x2 = 4Y The intersection of circle P and parabola C in the first quadrant is a, a (2,1) When x = 2, the slope of tangent k = y ′ = 12x = 1 The tangent is Y-1 = 1 × (X-2), that is, x-y-1 = 0 When x = 0, y = - 1, so Q (0, - 1) (10 minutes); ∵ the locus of the moving point m whose distance from P and Q is equal to 6 ∵ m is an ellipse with focus on the Y axis. Let its equation be x2b2 + y2a2 = 1 (a > b > 0) (12 points), then 2A = | MP | + | MQ | = 6, 2C = | PQ | = 2 The trajectory equation of M is x 28 + y 29 = 1 (14 points)
Expanding y = x * e ^ x into power series of X by indirect method
Y = x * ∑ (x ^ n / N!) = ∑ x ^ (n + 1) / N!, n from 0 to positive infinity