Seek the solution process of expanding the function f (x) = (2x + 1) / x ^ 2 + X-2 into a power series of (X-2)

Seek the solution process of expanding the function f (x) = (2x + 1) / x ^ 2 + X-2 into a power series of (X-2)

f(x)=(2x+1)/（x^2+x-2）=1/(x-1)+1/(x+2)=1/((x-2)-3)+1/((x-2)+4)
= 1/4 (1/(1+((x-2)/4))-1/3 (1/(1-((x-2)/3))
=1/4 Σ (-(x-2)/4) ^n - 1/3 Σ ((x-2)/3) ^n
= Σ 【(-1)^n/4^(n+1)-1/3^(n+1)】(x-2) ^n

Is the even function f (x) = ax square + 6x-2 defined on [1 + A, 2] an increasing function or a decreasing function on interval [1,2] Write down a reason

Because the definition field of even function is symmetrical about the origin
So 1 + A + 2 = 0
a=-3
So the opening is down
Because the axis of symmetry is x = - 6 / 2A = 1
So the function f (x) [1,2] is a subtractive function

It is known that the period of the function f (x) = asinwx + bcoswx + 1. (w.a.b is greater than 0) defined on R is π, F (π / 4) = root sign 3 + 1. And the maximum value of F (x) is 3. Find 1. Write the expression f (x) 2. Write the center, axis of symmetry, and axis of the heap 3. Explain how the image of F (x) is obtained from the image of function y = 2sin2x

f(X)=asinwx + bcoswx+1=√(a ²+ b ²) {[a/√(a ²+ b ²)] sinwx+[b/√(a ²+ b ²)] coswx}+1=√(a ²+ b ²) Sin (Wx + T) + 1 (where sin = B / √ (a)) ²+ b ²)) 1. If the period T = 2 π / w = π, then w = 2F (x)

It is known that the function f (x) is an odd function defined on (- 2,2). If f (x) = (2 ^ x) - 1 when x ∈ (0,2), the value of F (log2 1 / 3) is? Log2 1 / 3 means the logarithm of 3 based on 2

log2 1/3=-log2 30
∴f(log2 1/3)=f(-log2 3)
∵ odd function
∴f(log2 1/3)=f(-log2 3)=-f(log2 3)
∵ x ∈ (0,2), f (x) = (2 ^ x) - 1
∴f(log2 1/3)=-[2^(log2 3)-1]=-(3-1)=-2

Given that the function f (x) is an odd function defined on R, and x > 0, f (x) = log2 ^ (x + 1), then X

When X0
When x > 0, f (x) = log2 ^ (x + 1)
Therefore, f (- x) = log2 ^ (- x + 1)
If the function f (x) is an odd function, then there is f (- x) = - f (x)
So, when x

Function f (x) is an odd function defined on (- 2,2). When x is (0,2), f (x) = 2 ^ X-1, then f (x) = log2 (1 / 3)= The wrong number is to find the value of F (log2 (1 / 3)

I think your question is to ask the value of F (x) when x = log2 (1 / 3)
Because 1 / 4 < 1 / 3 < 1
-2F (x) is an odd function,
Then when x is (- 2,0), f (x) = - f (- x) = 1-2 ^ (- x)
Substitute x = log2 (1 / 3) into
At this time, f (x) = - 2