If n is a positive integer, then the value of [1 (- 1's 2nth power)] (n's Square-1) divided by 2 is 1. It's even. 2. It's odd 3. It's 0. 4. Odd and even numbers are possible

If n is a positive integer, then the value of [1 (- 1's 2nth power)] (n's Square-1) divided by 2 is 1. It's even. 2. It's odd 3. It's 0. 4. Odd and even numbers are possible

(1 (- 1's second n power))
[1/(-1)^2n]*(n^2-1)/2
=(n^2-1)/2
This result may be a fraction of it!

(2x + Y-1) to the second power, and (5 ^ 4 * 5 ^ 3-5 ^ 3 * 3 ^ 2 + 5 ^ 2 * 3) divided by 15

（2x+y-1）²
=(2x+y)²-2(2x+y)+1
=4x²+4xy+y²-4x-2y+1
（5^4×3^3-5^3×3^2+5^2×3)÷15
=5²×3×(5²×3²-5×3+1)÷15
=5×(225-15+1)
=5×211
=1055

Minus 10 of 23 minus 5 of 23 times 16 of 7

-10/23 - 5/23×16/7
= -5/23 × 2 - 5/23×16/7
= -5/23 × （2+16/7）
= -5/23 × 30/7
= -150/161

Using a 84 cm long wire to make a rectangular teaching aid, known as 9 cm long, 5 cm wide, how many cm high?
Hurry!

84 cm is the sum of edge length, so the sum of length, width and height is 84 △ 4 = 21 cm
So, the height is: 21-9-5 = 7 cm

12 divided by what is zero

12 divided by 0 is equal to 0. But 12 divided by 0 is also equal to 0, but 0 can't be a divisor. So 0 can be divided by 12, but 12 can't be divided by 0___ =0 then write 0

In the arithmetic sequence {an}, Sn = 5N ^ 2 + 3N is known. The general term formula of this sequence is obtained

When n = 1, A1 = S1 = 5 × 1 & # 178; + 3 × 1 = 8
When n ≥ 2, an = SN-S (n-1) = 5N & # 178; + 3N - [5 (n-1) &# 178; + 3 (n-1)] = 10n-2
When n = 1, A1 = 10 × 1-2 = 8, which also satisfies the general formula
The general formula of sequence {an} is an = 10n-2

3x plus 2 / 3 is equal to 5 / 6!
RT!

3x+2/3=5/6
3x=5/6-4/6=1/6
Then x = 1 / 18

(1) Xiao Yuan method 2x + y = 0 3x + 4Y = 2 (2) x-7y = 8 3x-8y = 10 (3) 4x + 3Y = 7 2x-3y = 5 (4) 2x-y = 5 3x + 4 and

It's the elimination method, right? It means to eliminate one unknown and then substitute the other into the formula
(1) 2X + y = 0 multiply by 3 to get 6x + 3Y = 0 and 3x + 4Y = 2 multiply by 2 to get 6x + 8y = 4 minus - 5Y = - 4
Y = 4 / 5 and then substitute it into any formula to get x = - 2 / 5

What are the integers of any three numbers from 1 to 9 that add up to 10

The maximum sum of three numbers from 1 to 9 is 7 + 8 + 9 = 24, so the maximum integral multiple of 10 is 20
The sum of 10 is 127, 136, 145 and 235
389, 479, 569 and 578 were added up to 20

X + y + Z + 2,4x + 2Y + Z + 2,9x + 3Y + Z = 6

x+y+z=2…… ①
4x+2y+z=2…… ②
9x+3y+z=6…… ③
②-①：
3x+y=0…… ④
③-②：
5x+y=4…… ⑤
⑤-④：
2x=4,∴x=2
Substituting 4:
6+y=0,∴y=-6
Substituting ①:
2-6+z=2,∴z=6
∴x=2,y=-6,z=6