In the binary linear equation 5x-3y = 16, if x and y are opposite to each other, the solution of the equation is ()

In the binary linear equation 5x-3y = 16, if x and y are opposite to each other, the solution of the equation is ()

Add the condition that X and y are opposite numbers
In fact, it's a system of linear equations of two variables
1、5x-3y=16
2、x+y=0
We get y = - x from formula 2 and substitute it into Formula 1
5x-3（-x）=16
8x=16
x=2
y=-x=-2
So the solution is x = 2, y = - 2

If the equation AX + 2bxy = 5x + 2Y about X and Y is a quadratic equation of two variables, then a and B must satisfy ()
A、a≠5,b=0 B、a≠5,b≠3/2 c、a≠-5,b≠3/2 D、a≠-5,b=0

If we choose a, if we want the formula to be a quadratic equation of two variables, then the term 2bxy must be 0, so B = 0. If we want to ensure that there are two unknowns, then a cannot be equal to 5, otherwise there will be only one unknowns. In conclusion, we choose a, a ≠ 5, B = 0

If the opposite number of X is - 8 and the absolute value of y = 5, find x + y=

13 or 3
Because x = - (- 8) = 8
There are two numbers with absolute value = 5
Take 5 as 8 + 5 = 13
-5 is 8-5 = 3

Given the function f (x) = 2x power - 2 negative x power sequence {an} satisfies f (log2an) = - 2n. (1) prove the sequence {an} decreasing sequence
(1) proving sequence {an} decreasing sequence

An > 0f (x) = 2 ^ X - 2 ^ (- x) = [2 ^ (2x) - 1] / 2 ^ XF (㏒ 2 an) = 2 ^ (㏒ 2 an) - 2 ^ (- ㏒ 2 an) = an - 1 / an = - 2n similarly, f (㏒ 2 an + 1) = (an + 1) - 1 / (an + 1) = - 2 (n + 1) subtracting (an + 1) - an + 1 / an - 1 / (an + 1) = - 2 [(an + 1)

In the sequence of negative 1,0,1 / 9,1 / 8 N squared n minus 2 Which item is. 0.08 in Chinese?

0.08=1/125,
1 / 125 = n of N squared parts minus 2, find n,
I calculated, it seems that there is no integer solution. Is the title copied correctly?

A sequence {an}, when n is odd, an = 5N + 1, when n is even, the sum of the first 2n terms of the sequence is

S(2n) = 5 * 1 +1 + 2^1
+ 5 * 3 + 1 + 2²
+ 5 * 5 + 1 + 2³
+ .
+ 5 * (2n-1) + 1 + 2^n
=n*1 + 5(1+3+5+ ...+ 2n-1) + (2+2² + 2³ + ...+ 2^n)
= n + 5*n² + 2 * (2^n-1)

It is known that the sum of the first n items of a sequence is n ^ 3, and the sum of the first n even items is (n ^ 2) * (4N + 3), then the sum of the first n odd items is

If the sum of the first n items is n ^ 3, then the sum of the first 2n items is (2n) ^ 3 = 8N ^ 3
So the sum of the first n odd terms is
8n^3-(n^2)*(4n+3)=4n^3-3n^2

Given the first n terms of the sequence {an} and the n-th power-1 of Sn = a (a is a non-zero constant), then the sequence {an} is an arithmetic sequence
How to prove it
Sn=(A^n)-1

n≥
Sn=(A^n)-1
Sn-1=A^(n-1)-1
An=Sn-Sn-1=A^n-1-[A^(n-1)-1]
=A^n-A^(n-1)
n=1
S1=A-1
According to an
An=A^1-A^0=A-1
So {an} is an arithmetic sequence, and the general formula an = a ^ n-a ^ (n-1)

Given that the first n terms and Sn of sequence {an} satisfy: SN = A / A-1 (an-1) (a is a constant, and a is not equal to 0, a is not equal to 1)
Given that the first n terms and Sn of sequence {an} satisfy: SN = A / A-1 (an-1) (a is constant, and a is not equal to 0, a is not equal to 1); (1) find the general term formula of sequence {an}; (2) let BN = 2Sn / an + 1, if {BN} is an equal ratio sequence, find the value of a; (3) in the case of (2), let CN = 1 / 1 + an + 1 / 1-an + 1, the sum of the first n terms of sequence {CN} be TN, and prove that TN is greater than 2N-1 / 3

1) When n = 1, S1 = A1 = A / (A-1) (A1-1), A1 = a, when n ≥ 2, an = SN-S (n-1) = A / (A-1) (an-a (n-1)) shifts to an = a * a (n-1), that is, an / a (n-1) = a, so {an} is an equal ratio sequence with a common ratio, so the general formula of {an} is an = a ^ N 2. From 1), Sn = a (a ^ n-1) / (A-1) can be obtained

Given the first n terms of sequence {an} and Sn = a ^ n-1 (a is not a constant of zero), what kind of sequence is sequence {an}
For example, it must be an arithmetic sequence, it must be an arithmetic sequence, it must be an arithmetic sequence, it must be an arithmetic sequence, it must be an arithmetic sequence, it must be an arithmetic sequence, it must be an arithmetic sequence, it must be an arithmetic sequence, it must be an arithmetic sequence, it must be an arithmetic sequence, it must be an arithmetic sequence, it must be an arithmetic sequence, it must be an arithmetic sequence, it must be an arithmetic

a1=s1=a^1-1=a-1
an=Sn-S(n-1)=a^n-1-(a^(n-1)-1)=(a-1)*a^(n-1)
When a = 1, an = 0, an-a (n-1) = 0, so {an} is an arithmetic sequence with 0 as the head and 0 as the tolerance
When a! = 1
An/A(n-1)=(a-1)*a^(n-1)/((a-1)*a^(n-1-1)=a
So when a! = 1, {an} is an equal ratio sequence with A-1 as the head and a as the common ratio