When x is equal to half, the sum of the square of X, which is one fifth of the algebraic expression, and 1

When x is equal to half, the sum of the square of X, which is one fifth of the algebraic expression, and 1

1 / 5 * (square of X + 1) = 1 / 5 * (1 / 4 + 1) = 1 / 4 = 0.25

If M (3b + a) = - 3 / 2 ab - 1 / 2 a squared

M=(-3/2ab-1/2a²)/(3b+a)
=-1/2a(3b+a)/(3b+a)
=-1/2a

How much does (the 2000th power of 10 minus 1) * (the 2000th power of 10 minus 1) equal

You can write the result directly to the power of 10 to the power of 4000 ~ no problem

2n + 1 power of 2 + 2n power of 2 = 48 to find the value of n

Is the 2n plus 1 power of 2 twice the 2n power of 2
Then the problem is reduced to 3 times the 2n power of 2 = 48
That is, the 2n power of 2 is equal to 16
Obviously, the fourth power of 2 is 16
So n = 2

In 1,0,2,5 / 6,11,15,95, integers have (), odd numbers have (), prime numbers have (), and combined numbers have ()

Integers have 1,0,2,11,15,95
Odd numbers have 1, 11, 15, 95
Prime numbers have two, eleven
Total 15,95

There is a 2n + 1-bit integer (n is an integer, n is greater than or equal to 1) 22... 23 (n-bit 2) 11... 1 (n-bit 1). Is it a prime number or a composite number?

Let the original number be mm = 2 × 10 ^ 2n + 2 × 10 ^ (2n-1) + +2×10^(n+1)+3×10^n+10^(n-1)+10^(n-2)+…… +Note that 3 × 10 ^ n = 2 × 10 ^ n + 10 ^ n, then M = 2 × 10 ^ 2n + 2 × 10 ^ n (2n-1) + +2×10^(n+1)+2×10^n+10^n+10^(n-1)+10^(n-2)+…… +1...

11... 1211... 1
There are n ones in front of 2 and N ones after 2. N is any natural number

11...1211...1
=11...1100...0+111...1
=111...1*（100...0+1）
=111...1*100...1）
Total number

It is proved that if n is a composite number, then 2 ^ n-1 is also a composite number

If n is a composite number, n = PQ, P ≥ 2, Q ≥ 2
2^n-1 = 2^(pq)-1
= (2^p-1)(2^q+2^(q-1)+...+2+1)
2^p-1 > 2,2^q+2^(q-1)+...+2+1 >2
2 ^ n-1 is also a composite number

Let n be a composite number 11 (n) is a composite number or a prime number, and prove

A: this number is a composite number. Prove: since n is a composite number, then n = a × B, where a and B are two divisors of n that are not 1. Then the original number can be divided into a segments, each segment is 11 11 (b). Write these numbers in the form of sum and arrange them from small to large as follows: 11 11 (n) = 11 11 (b) + 11 11 (b

For each positive integer n, the number 11 1211… Is 1 a sum? Why?
11… 1211… 1 (there are n)! Please help me!

It's all a sum
121 = 11 * 11
11211 = 111 * 101
1112111 = 1111 * 1001
……
This is the law. If the original number 2 has n ones each, it can be decomposed into the product of 11.. 1 (n + 1 ones) and 100.. 01 (n-1 ones)