10/3+（-11/4）-（-5/6）+（-7/12）

10/3+（-11/4）-（-5/6）+（-7/12）

10/3+（-11/4）-（-5/6）+（-7/12）
=10/3-11/4+5/6-7/12
=40/12-33/12+10/12-7/12
=10/12
=5/6;
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The "H operation" of positive integer n is: 1. When n is odd, H = 3N + 13; 2. When n is even, H = n * 0.5 * 0.5 * (where h is an odd number). For example, the result of the number 3 after one "H operation" is 22, the result after two "H operations" is 11, and the result after three "H operations" is 46
(1) After 257 times of "H operation", the result is obtained;
(2) If the result of "H operation" 2 is always constant a, find the value of A

257
Once = 3 * 257 + 13 = 784
Twice = 784 * 0.5 * 0.5 * 0.5 * 0.5 = 49
3 times = 3 * 49 + 13 = 160
4 times = 160 * 0.5 * 0.5 * 0.5 * 0.5 * 0.5 * 0.5 = 5
5 times = 3 * 5 + 13 = 28
6 times = 28 * 0.5 * 0.5 = 7
7 times = 3 * 7 + 13 = 34
8 times = 34 * 0.5 = 17
9 times = 3 * 17 + 13 = 64
10 times = 64 * 0.5 * 0.5 * 0.5 * 0.5 * 0.5 * 0.5 * 0.5 = 1
11 times = 3 * 1 + 13 = 16
12 times = 16 * 0.5 * 0.5 * 0.5 * 0.5 = 1 = the 10th time
So start from the 10th
An even number of times equals one
Odd times equal to 16
257 is odd
So 257th is 16
If a cycle occurs after several times of "H" operation on a positive integer, then the result of "H" operation is always a, then a must be an odd number
Then, the result of H operation on a * 3 + 13 is even, and then "H operation" is performed on a * 3 + 13
A * 3 + 13 times 1 / (2 ^ k) is still a
So (a * 3 + 13) * 1 / 2 ^ k = a
That is a * 3 + 13 = a * 2 ^ K
That is, a (2 ^ K-3) = 13 = 1 * 13
Because a is a positive integer
SO 2 ^ K-3 = 1 or 2 ^ K-3 = 13
The solution is k = 2 or K = 4
When k = 2, a = 13; when k = 4, a = 1, so a is 1 or 13

Prove by mathematical induction: 1 + 3 + 5 + +（2n-1）=n2．

It is proved that: (1) when n = 1, left = 1, right = 1, left = right. (2) suppose n = k, the equation holds, that is, 1 + 3 + 5 + +(2k-1) = K2 when n = K + 1, the left side of the equation = 1 + 3 + 5 + +(2k-1) + (2k + 1) = K2 + (2k + 1) = (K + 1) 2 +(2n-1) = N2 holds for any positive integer

Number______ The sum of - 512 equals - 78

-78 - (- 512) = - 78 + 512 = - 1124

What is split term elimination?

（1）1/[n(n+1)]=（1/n）- [1/(n+1)]
（2）1/[(2n-1)(2n+1)]=1/2[1/(2n-1)-1/(2n+1)]
（3）1/[n(n+1)(n+2)]=1/2{1/[n(n+1)]-1/[(n+1)(n+2)]}
（4）1/(√a+√b)=[1/(a-b)](√a-√b)
（5） n·n!=(n+1)!-n!
（6）1/[n(n+k)]=1/k[1/n-1/(n
+k) Examples: 1 / (1 × 4) + 1 / (4 × 7) + 1 / (7 × 10) + +1 / (91 × 94) each fraction is expanded into two fractions by using split term formula
*[（1-1/4）+（1/4-1/7）+（1/7-1/10）+…… +（1/91-1/94）]=1/3*(1-1/94)=31/94

Function f (x) = 2x, find the image of function y = | f (x + 1) - 1 | and write the monotone interval

f(x)=(1-x)/(1+x)
=[2-(1+x)]/(1+x)
=[2/(1+x)]-1
That is, y = 2 / X moves left by 1 and down by 1
Monotone decreasing interval (negative infinity, - 1), (- 1, positive infinity)

What is the greatest common factor and the least common multiple of 12, 18 and 30

12=2*3*2
18=2*3*3
30=2*3*5
The greatest common factor is the overlapping part 2 * 3 = 6
The least common multiple is the non overlapping part of the greatest common factor * 6 * 2 * 3 * 5 = 180
The method is so universal

Given that a is greater than 0b is greater than 01 / A + 2 / b = 1, find the minimum value of ab

Sum: SN = 1 + 2x + 3x ^ 3 +. + NX ^ n-1
1 + 2x + 3x ^ 3 the third power of 3x... Not the square... But thank you for your answer

Sn = 1 + 2x + 3x ^ 2 +. + NX ^ (n-1) xsn = x + 2x ^ 2 + 3x ^ 3 +... + NX ^ n the following formula minus the upper formula (x-1) Sn = - 1-x-x ^ 2 -... - NX ^ (n-1) + NX ^ n (x-1) Sn = NX ^ n - (1-x ^ n) / (1-x) Sn = NX ^ n + (1-x ^ n) / (x-1) ^ 2 is the square. If it's 3x ^ 3, how can it become NX ^ (n-1)? ^ n

Calculate 1 + 2-3-4 + 5 + 6-7-8 + 9 + +26-27-28-29
a.1 b.-1 c.-28 d.-57
I'm - 58, too

1+2-3-4+5+6-7-8+9+… +26-27-28-29
= (1+2-3-4)+(5+6-7-8)+… +(25+26-27-28)-29
= (-4)×7 -29
= -28 -29
= -57
So choose D