Evaluation: 1 + X + X & # 178; + X & # 179; + X & # 8308; + ······· + X & # 8319; n = 2011 and 1 + X + X & # 178; + X & # 179; = 0

Evaluation: 1 + X + X & # 178; + X & # 179; + X & # 8308; + ······· + X & # 8319; n = 2011 and 1 + X + X & # 178; + X & # 179; = 0

1+x+x²+x³+x⁴+·······+xⁿ
=1+x(1+x+x²+x³)+x⁴(1+x+x²+x³)+...+x^2008(1+x+x²+x³)
=1

x⁴2x⁴y+x⁴y²-2x²+y²-2x²y²+2y+1
Factorization

In the second item, a symbol is missing in the second item, and the guess is that the conjguess should be the minus sign. The conjconjconjconjconjconjconjconjguess is the minus sign. The conjconjconjconjconjconjconjconjguess is the minus sign, which is: X \3583308; - x \35\35\35\?; - 2x \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\;) = (X & # 178; - 1) &# 178; Y & # 178

The image of derivative function f '(x) of function f (x0 = xcosx) on interval [- π, π]

f '（x）=cosx-xsinx
The image looks like this

If 2x-y = 3,2y-z = 4,2z-x = 5, then x + y + Z =?

A: x + y + Z = 12
2x-y+(2y-z）+（2z-x)=3+4+5 =12
2x-y+2y-z+2z-x=12
x+y+z=12
It's interesting to do math problems. Think more·

In Cartesian coordinate system, given points a (4,0), B (0,3), if a right triangle is congruent with RT △ ABO, and they have a common edge, please write out the unknown vertex coordinates of the triangle (do not need to write the calculation process). (hint: consider the three cases of Ao, Bo, AB being the common edge respectively)

As shown in the figure, if AB is the common edge, there are three answers (72259625), (4,3), (2825, - 2125); if Bo is the common edge, there are two answers (- 4,3) and (- 4,0); if Ao is the common edge, there are two answers (0, - 3) and (4, - 3)

Solution equation: 2 (x-3) ≤ 9 - (x + 12)

2x-6

Proof: let vector group A1, A2, A3 be linearly independent, and prove that vector group A1 + 2A2, A2 + 2A3, A3 + 2A1 are linearly independent

Let K1, K2, K3 make K1 (a1 + 2A2) + K2 (A2 + 2A3) + K3 (A3 + 2A1) = 0 (K1 + 2K3) a1 + (2K1 + K2) A2 + (2k2 + K3) A3 = 0a1, A2, A3 linearly independent, so K1 + 2K3 = 02k1 + K2 = 02k2 + K3 = 0 solution: K1 = K2 = K3 = 0, so vector group A1 + 2A2, A2 + 2A3, A3 + 2A1 linearly independent

If an equation has an irrational root in the interval (0,2) and the accuracy of the approximate solution is not more than 1 / 2 of the nth power, then at least
How many times does interval (0,2) equal?
Why?

Half at a time
I think it's irrational
Let the error be d
The first time is d ＜ (2-0) / 2 = 1
The second time d < 1 / 2
The third time D ＜ (1 / 2) ^ 2
The fourth D ＜ (1 / 2) ^ 3
That is the nth time
d＜(1/2)^(n-1)
The accuracy is not more than 1 / 2 to the nth power
That is, the N + 1st time
In this case, d < 1 / 2) ^ n
It's already quite detailed. Hope to adopt it``````
Bonus points``````

Let Sn be the sum of the first n terms of the arithmetic sequence {an}, if a5a3 = 59, then s9s5 = ()
A. 1B. -1C. 2D. 12

Let the first term of the arithmetic sequence {an} be A1. From the properties of the arithmetic sequence, we can get a1 + A9 = 2a5, a1 + A5 = 2A3, ｜ s9s5 = a1 + a92 × 9a1 + a52 × 5 = 9a55, A3 = 95 × 59 = 1, so we choose a

Let f (x) = 2x3-3 (A-1) x2 + 1, where a ≥ 1. (I) find the monotone interval of F (x); (II) discuss the extremum of F (x)

From the known f '(x) = 6x [x - (A-1)], Let f' (x) = 0, the solution is X1 = 0, X2 = A-1. (I) when a = 1, f '(x) = 6x2, f (x) increases monotonically on (- ∞, + ∞). When a > 1, f' (x) = 6x [x - (A-1)], f '(x), f (x) changes with X as follows: from the above table, we can see that the function f (x) increases monotonically on (- ∞, 0); when a > 1, f' (x) = 6x [x - (A-1)], f '(x), f (x) increases monotonically on (0, A-1) It is known from (I) that when a = 1, the function f (x) has no extremum. When a > 1, the function f (x) has a maximum at x = 0 and a minimum at 450, 1 - (A-1) 3