Given that the absolute value of X + 8 and the absolute value of Y-2 / 5 are opposite to each other, what is the value of X + y

Given that the absolute value of X + 8 and the absolute value of Y-2 / 5 are opposite to each other, what is the value of X + y

The absolute value of X + 8 is opposite to that of Y-2 / 5
|x+8|=-|y-2/5|
|x+8|+|y-2/5|=0
x=-8 y=2/5
x+y=-8+2/5
=-38/5

If the absolute value of X + 8 and the absolute value of y-4 are opposite to each other, the value of X-Y is obtained

Solution
/ x + 8 / and / y-4 / are opposite numbers
∴/x+8/+/y-4/=0
∵/x+8/≥0
/y-4/≥0
∴x+8=0,y-4=0
∴x=-8,y=4
∴x-y=-8-4=-12

The absolute value of a / (A-A ＾ 2) = 1 / A-1 to find the value range of A

A ≠ 1 and a ≠ 0

It is known that the sum of the first n terms of the sequence {an} is the nth power of Sn = (- 1) * n, an=

an=Sn-Sn-1=(-1)^n*n-(-1)^(n-1)*(n-1)=(-1)^n(2n-1)

Sn = 1 + 2 + 4 +... + 2

Sn = 1 × (n + 1 power of 1-2) / (1-2) = n + 1 power of 2-1

Equal ratio sequence: 1 / 4 + 1 / 8 + '+ 1 / 2 n power, how to find Sn?

The dislocation subtraction
Let this sequence be SN
Then (1 / 2) Sn = 1 / 8 + 1 / 16 + '+ 1 / 2 ^ n + 1 / 2 ^ (n + 1)
Then the two equations are subtracted
Will you?
I wrote it completely

Known: 26 = A2 = 4b, find the value of a + B

Therefore, a + B = 8 + 3 = 11 or a + B = - 8 + 3 = - 5

The last digit of (2 + 1) * (2 · 2 + 1) * (2 · 4) * (2 · 8 + 1) * (2 · 16 + 1) * (2 · 64 + 1) + 1 is

=(2-1)(2+1)*(2¤2+1)*(2¤4)*(2¤8+1)*(2¤16+1)*(2¤64+1)+1
=2¤128-1+1
=2¤128
2¤1=2
2¤2=4
2¤3=8
2¤4=16
The period at the end of the cycle is 4 128 / 4 = 32
Should end with 6

How to calculate 1 / 2 + 1 / 4 + 1 / 8 + 1 / 16 +... + 1 / (n power of 2)

1/2+1/4+1/8+1/16+...+1/2^n
=1/2*(1-(1/2)^n)*/(1-1/2)
=1-1/2^n

When n is a positive integer, the nth power of 37.5 plus the nth power of 26.5 is also a positive integer
Online help looking forward to the answer. The first time to score! Every 5 minutes I see if there is an answer!
Not only the answer, but also the proof! Friends really answer very quickly, but also have to prove Oh! It's good to say 1, but it's wrong to use binomial. Because except for 27 ^ n, the following items are not necessarily integers.
N power of 37.5 plus N power of 26.5
=(75/2)^n+(53/2)^n
=(75^n+53^n)/2^n
Is an integer, set to a, then
75 ^ n this proof has a play, but it's not finished? Everyone speed point, I see the detailed answer to lose points immediately!!

It's good to be 1, but it's wrong to prove it with binomial, because except 27 ^ n, the following items are not necessarily integers
N power of 37.5 plus N power of 26.5
=(75/2)^n+(53/2)^n
=(75^n+53^n)/2^n
=[(64+11)^n+(64-11)^n]/2^n
Using binomial theorem, we can know that after expansion, when the power of 11 is odd, it will cancel out; when the power of 11 is even, it will add up
Finally, it can be concluded that when n is odd, it always satisfies the meaning of the problem