1 / (1 + x) + 2 / (1 + x ^ 2) + 4 / (1 + x ^ 4) + 8 / (1 + x ^ 8) = 0 find 16 / (1-x ^ 16) - 1 / (1-x)

1 / (1 + x) + 2 / (1 + x ^ 2) + 4 / (1 + x ^ 4) + 8 / (1 + x ^ 8) = 0 find 16 / (1-x ^ 16) - 1 / (1-x)

16 / (1-x ^ 16) - 1 / (1-x) = 16 / [(1-x ^ 8) (1 + x ^ 8)] - 1 / (1-x) = 8 / (1-x ^ 8) + 8 / (1 + x ^ 8) - 1 / (1-x) in the same way, 8 / (1-x ^ 8) is further decomposed into 16 / (1-x ^ 16) - 1 / (1-x) = 16 / [(1-x ^ 8) (1 + x ^ 8)] - 1 / (1-x) = 1 / (1 + x) + 2 / (1 + x ^ 2) + 4 / (1 + x ^ 4) + 8 / (1 + x ^ 8) +

(x-1) (x + 1) = x ^ 2-1, (x-1) (x ^ 2 + X + 1) = x ^ 3-1. Given x ^ 4 + x ^ 3 + x ^ 2 + X + 1 = 0, find x ^ 2010

Because (x-1) (x ^ 4 + x ^ 3 + x ^ 2 + X + 1) = x ^ 5 - 1 = 0
So x ^ 5 = 1
So x ^ 2010 = x ^ (5 × 402) = (x ^ 5) ^ 402 = 1 ^ 402 = 1
I hope it works

5/x^2+x - 1/x^2-x=0

5/x(x+1)-1/x(x-1)=0
Don't multiply x (x + 1) (x-1)
5(x-1)-(x+1)=0
5x-5-x-1=0
x=3/2

Given the function f (x) = 2 + log3x, X ∈ [1,9], find the maximum value of y = [f (x)] & sup2; + F (X & sup2;), and the value of X when y gets the maximum value

Let log3x = t, then t ∈ [0,2]
f(x²)=2+2t
So y = [f (x)] & sup2; + F (X & sup2;) = T ^ 2 + 6T + 4 = (T + 3) ^ 2-5
And t ∈ [0,2], so the maximum value of Y is obtained when t = 2
In this case, x = 9, y max = 20

Find the monotonicity and monotone interval of the function y (x) = x & # 179; - X & # 178; - X,

f’（x）=3x^2-2x-1=(3x+1)(x-1)
Let f '(x) > 0 get x > 1 or X

What is the original number of a denominator with 6 numerators and 4 numerators
No equations!

The denominator is a multiple of 6, and the denominator is 6 times the numerator minus 6
The denominator is four times that of the numerator plus four,
The molecular weight is [6 + 4 * 4] / (6-4) = 11
The denominator is 60
So it's 11 / 60

General solution of differential equation (x-1) y '= y (1 + 2XY)

General solution of differential equation (x-1) y '= y (1 + 2XY)
In order to make up the differential, divide both sides by Y & #178;, we get
y'(x-1)/y²=(1+2xy)/y
(dy/dx)(x-1)/y²=(1+2xy)/y
(x-1)dy/y²=(1+2xy)dx/y
[(x/y²)-(1/y²)]dy=[(1/y)+2x]dx
(x/y²)dy-(1/y²)dy=(1/y)dx+2xdx
-(1/y²)dy-2xdx=(1/y)dx-(x/y²)dy
Because - (1 / Y & # 178;) dy = 1 / y, 2xdx = x & # 178;,
The differential of [x * (1 / y)] = DX * 1 / y + X (- 1 / Y & # 178;) dy = (1 / y) DX - (x / Y & # 178;) dy,
So, there is 1 / Y-X & # 178; = x / y + C
So the general solution is: X & # 178; + X / Y-1 / y + C = 0

3 / 8, - 5 / 24,7 / 48, - 9 / 80... A general formula is

The general formula is an = [(- 1) ^ (n + 1)] * (2n + 1) / [4N (n + 1)]

Fill in the box with 123456789 to make the formula true
one
{< >< >+< >-< >=< >
{< >*< >=< >< >
★★★
Don't overdo it. Just fill in the box

12+3-7=8
6*9=54

Given that ABCD is a parallelogram, and a (4,1,3), B (2, - 5,1), C (3,7, - 5), then the coordinates of point D are______ ．

By bisecting the two diagonals of the parallelogram, the sum of the coordinates of a and C is equal to that of B and D. let d be (x, y, z), then 4 + 3 = 2 + X1 + 7 = − 5 + Y3 − 5 = 1 + Z. the solution is: x = 5Y = 13z = − 3, so the answer is: (5, 13, - 3)