Sum: SN = 1-3x + 5x ^ 2-7x ^ 3 +. + (2n + 1) (- x) ^ n (n belongs to n *)

Sum: SN = 1-3x + 5x ^ 2-7x ^ 3 +. + (2n + 1) (- x) ^ n (n belongs to n *)

Sum: SN = 1 + 3x + 5x + 7x +... + (2n-1) x ^ (n-1) thank you,

sn=1+3x+5x^2+7x^3+...+（2n－1）x^n-1 （1） xsn= x+3x^2+5x^3+...+(2n-3)x^n-1+（2n－1）x^n （2） （1）－（2）:(1-x)sn=(1+2x+2x^2+2x^3+...+2x^n-1)-（2n－1）x^n =2[(x^n)-1]/(x-1)-1-(2n-1)x^n Sn=2[(x^n)-1)...

How many times and terms is 3x & # 178; Y-X & # 179; Y & # 179; - 9xy & # 178; + 1?

It's six times four

{2} Now you can use those methods to get the equations {5 [x + y] - 3 [X-Y] = 16 {3 [x + y] - 5 {X-Y] = 0 2010-12-17 18:49

5 [x + y] - 3 [X-Y] = 16 open brackets: 5x + 5Y - (3x-3y) = 165x + 5y-3x + 3Y = 162x + 8y = 16 equation 13 [x + y] - 5 {X-Y] = 03x + 3Y - (5x-5y) = 03x + 3y-5x + 5Y = 08y-2x = 0 equation 2 equation 1 + 2 get 16y = 16 y

The reduction of (A's second power + 5a) - 2 (half a's second power-4a)
There are also the quadratic power of a = x-5x, the quadratic power of B = x-10x + 3, the value of 2a-b is known as - M + 2n = 2, and the quadratic power of 5 (m-2n) + 6n-3m-3

=a^2+5a-a^2+2a
=7a
2A-B=2(x^2-5x)-(x^2-10x+3)
=2x^2-10x-x^2+10x-3
=x^2-3
5(m-2n)^2+6n-3m-3
=5(-m+2n)^2+3(2n-m)-3
=5x2^2+3x2-3
=26-3
=23

I know how to move the term of linear equation of one variable. I also know the sign change. That is, where to move it
For example. 3x-5 + 6 = 5x + 7, I said if you want to move the item. 5x has passed and changed the sign. - 5 and + 6 move over. Where do you want to put them? Let me give you an example. Explain what I mean, specifically. Move over and put them in

3x-5+6＝5x+7
2x=-6
x=-3
For linear equation of one variable, the constant term is moved to the right and the unknown term to the left

How to understand the derivation of this implicit function
Problem siny + cosx = 1 solution: both sides of the equation at the same time to the independent variable x derivation cosy * y ^ - SiNx = 0, why multiply y ^ ah? Also, ah, why the X + y times of E after X derivation will be equal to the X + y times of e multiplied by bracket 1 + y ^ ah, really don't understand ah, I know the big brother to help explain, thank you
And why does the SiNx power of y = x have such an equality as iny = sinxinx after taking logarithm on both sides of the equation? What is the intermediate process?

If you think of Y as a function of X, y = f (x), it's easy to understand
siny+cosx=1
y'cosy-sinx=0
y'=sinx/cosy
In this case, is it more understandable
sinf(x)+cosx=1
f'(x)cosf(x)-sinx=0
[e^(x+y)]'=e^(x+y)*(x'+y')=(1+y')e^(x+y)

Factorization of 4A (B-A) - B ^ 2; a ^ 2 (B + 1) - B ^ 2 (a + 1); a ^ 2 (a-b) 2 (m-n) + B ^ 2 (B-A) ^ 2 (n-m)

Y is equal to the absolute value 2x + 1, X belongs to [- 1,1]

Y is equal to the absolute value 2x + 1, X belongs to [- 1,1]
For 2x + 1;
-2+1≤2x+1≤2+1；
That is - 1 ≤ 2x + 1 ≤ 3;
So the range is [0,3]
I'm very glad to answer your questions. Skyhunter 002 will answer your questions
If you don't understand this question, you can ask,

If the function f (x) = x2-1 is known, then the zero of F (x-1) is zero______ ．

When ∵ function f (x) = x2-1, ∵ f (x) = x2-1 = 0, x = 1 or - 1, the comparison between function f (x-1) and f (x) is that the image of function f (x) shifts one unit to the right, and the zero point of ∵ f (x-1) is 1 + 1 = 2, - 1 + 1 = 0, so the answer is: 0, 2