N times the nth power of Q, n tends to infinity, 0

N times the nth power of Q, n tends to infinity, 0

How could it be 1
1 / (Q ^ n) is a higher order infinitesimal of 1 / n
The answer is 0

The second power of X + X + the third power of X + the fourth power of X + How to calculate the nth power of + X?

Is the sum of equal ratio sequence
N times of the first term x and the last term x
When x is not equal to 0
Sn = (n + 1 power of X + x) / (1-x)
If x = 0
Then Sn = 0

To make the 2n power of polynomial 3x + M-1 of 6x be a quadratic trinomial of X (M and N are positive integers), find the values of M and n

∵ the polynomial 3x ^ 2n + 6x ^ M-1 is a quadratic trinomial of X
ν 2n = 2 and M = 1 or 2n = 1 and M = 2
The solution is n = 1, M = 1 or n = 1 / 4, M = 2
And ∵ m, n are positive integers
ν n = 1 / 4, M = 2
∴n=1,m=1

Given that the quadratic power of | a + 5 | + (B-4) is equal to 0, what is the 2013 power of (a + b)?

|The quadratic power of a + 5 | + (B-4) is equal to 0
So a + 5 = 0, B-4 = 0
a=-5,b=4
So a + B = - 1
Therefore, the 2013 power of the original formula = (- 1) = - 1

Then the power of (n) to the power of n is equal to the power of 4x=

The 2n power of (XY) is the 2n power of X * the 2n power of Y
=(to the nth power of x) 2 * (to the nth power of Y)
=5²*4²
=25*16
=400

The square of X + X-1 = 0, then what is the cubic power of X + the square of 2x - 7 ditto

x^2+x-1=0
x^2+x=1
x^3+2x^2-7
=x(x^2+x)+x^2-7
=x+x^2-7
=1-7
=-6

Let x = YY be a function of X, then dy=______ .

x=yy
lnx=ylny
One
xdx＝lnydy+y×1
ydy＝(1+lny)dy
dy＝1
x(1+lny)dx．

We know that the set a = y|y = (1 / 4) to the X + 1 power of-3 (1 / 2), and X belongs to (- 1,2) B = x | x-m square | ≥ 1 / 4, Proposition p x belongs to a, proposition q x belongs to B, and proposition p is a sufficient condition of proposition Q. the value range of real number m is obtained

A:
y=(1/4)^x-3*(1/2)^(x+1)
=[(1/2)^x]^2-(3/2)*(1/2)^x
=[(1/2)^x-3/4]^2-9/16
Because: - 1

Let u = R be a complete set, a = {X / Log1 / 2 [(3-x)] ≥ - 2}, B = {X / 2 to the power of - 2x + 1

Analysis:
The inequality log (1 / 2) (3-x) ≥ - 2 in set a can be reduced to log (1 / 2) (3-x) ≥ log (1 / 2) 4
The solution is: 3-x > 0 and 3-x ≤ 4
That is - 1 ≤ X

The derivative of y = ln (1 + e to the power of x)

y'=[1/(1＋e^x)]×(e^x)'=(e^x)/(1＋e^x)