Compare the size of 3,4,3 root 50 I figured it out ∵3³=27， 3 = cubic root 27 ∵4³=64， 4 = cubic root 64 3 ＜ triple root 50 ＜ 4

Compare the size of 3,4,3 root 50 I figured it out ∵3³=27， 3 = cubic root 27 ∵4³=64， 4 = cubic root 64 3 ＜ triple root 50 ＜ 4

Because the third power of 3 is equal to 27, and the third power of 4 is equal to 64, so the root of 3 is 50

(1) (7 + 4 √ 3) ^ 1 / 2-27 ^ 1 / 6 + 16 ^ 3 / 4-2 * (8 ^ - 2 / 3) ^ - 1 + √ 2 quintic root * (4 ^ - 2 / 5) ^ - 1 (2) (1 / 3) ^ - 1 / 2 + √ 3 * (√

1、
The original formula = [(2 + √ 3) & sup2;] ^ (1 / 2) - (3 & sup3;) ^ (1 / 6) + (2 ^ 4) ^ 3 / 4-2 * 8 ^ (2 / 3) + 2 ^ (1 / 5) * 2 ^ (4 / 5)
=2+√3-3^1/2+2^3-2*2^2+2^(1/5+4/5)
=2+√3-√3+8-8+2
=4
2、
Incomplete

30-29 + 28-27 +. + 2-1 =? Simple operation,

An arithmetic sequence {AK} has n terms, n is odd, what is 132 for all odd terms, 120 for all even terms,
(1) , find n
(2) If an = 2, find the general term AK

（1）
Let n = 2m + 1, m be a nonnegative integer, d be the tolerance of AK
be
132=（a1+a1+2md）（m+1）/2=(a1+md)(m+1) …… First form
120=[a1+d+a1+(2m-1)d]m/2=(a1+md)m …… second form
12 = (a1 + MD) The third form
The second formula plus the first formula is 252 = (a1 + MD) (2m + 1) The fourth form
The fourth formula is divided by the third formula: n = 2m + 1 = 252 / 12 = 21
（2）
From above we know that M = 10;
132=（an+an-2md）（m+1）/2=(an-10d)(10+1)
All d = - 1, negative 1
a1=22
So AK = 23-k
The solution is over

How to change a 100 yuan note into a 1 yuan, 2 yuan, 5 yuan, 10 yuan note? At least one of each note. How many ways do you have

There are 16 kinds. (the number in brackets is used to replace the number below)
1、 When 5 yuan has 2, 10 yuan can have (7. 5. 3. 1), 20 yuan can have (1. 2. 3. 4). 10 yuan and 20 yuan are corresponding, that is, when 10 yuan is 7, 20 yuan is 1. The following is the same
2、 When 5 yuan has 4, 10 yuan can have (6. 4. 2), 20 yuan can have (1. 2. 3)
3、 When 5 yuan has 6, 10 yuan can have (5. 3. 1), 20 yuan can have (1. 2. 3)
4、 When 5 yuan has 8, 10 yuan can have (4. 2), 20 yuan can have (1. 2)
5、 When 5 yuan has 10, 10 yuan can have (3. 1), 20 yuan can have (1. 2)
6、 When 5 yuan has 12, 10 yuan has 2, 20 yuan has 1
7、 When 5 yuan has 14, 10 yuan has 1, 20 yuan has 1

1 / 20.4 37.5%: 4 / 11 4 / 15 () (fill in the score) () (fill in the hundred distribution)

1 / 2 2 / 5 3 / 8 4 / 11 4 / 15 I think there is something wrong with the number

It is known that the quadratic function f (x) = ax ^ 2 + BX satisfies the following conditions: ① f (0) = f (1) ② the minimum value of F (x) is - 1 / 8, let the first N-term product of sequence {an} be TN, and TN = (4 / 5) ^ f (n), and find the general term formula of sequence {an}

Under the condition 1: F (0) = 0; f (1) = a + B = 0; the symmetry axis of quadratic function is 1 / 2
Let 2 be a > 0, f (1 / 2) = - 1 / 8 = 1A / 4 = 1B / 2, that is, 2A + 4B = - 1,
A = 1 / 2, B = - 1 / 2
Then the function expression is: F (x) = (x ^ 2) / 2-x / 2
Because the product of the first n terms of the sequence {an} is TN = (4 / 5) ^ f (n)
Then the product of the first n terms of the sequence {a (n-1)} is t (n-1) = (4 / 5) ^ f (n-1)
So when n > 1, an = TN / T (n-1) = (4 / 5) ^ (n-1)
When n = 1, A1 = T1 = 1 satisfies the above formula
So to sum up, the sequence an = (4 / 5) ^ (n-1)

Sqrt (1 + ((- 2 * x) / (1-x * x)) * ((- 2 * x) / (1-x * x))) integral

First complete square 1-x-x ^ 2 = 5 / 4 - (x ^ 2 + X + 1 / 4) = (sqrt (5) / 2) ^ 2 - ((1 + 2x) / 2) ^ 2

Find the rules: what's after 1,1,2,3,4,6,9,19?

1,1,2,3,4,6,9,13,19 this is the original question. The next 28 1 + 3 = 4 2 + 4 = 6 3 + 6 = 9 4 + 9 = 13 6 + 13 = 19

The sum of a nonzero natural number multiplied by a false fraction is greater than the original number

It is wrong that the product of a nonzero natural number multiplied by a false fraction is greater than the original number. Because the false fraction may be equal to 1, then the product is equal to the original number