Use the natural number a to get rid of the fact that the quotient of 2008 is 44. What is the natural number and what is the remainder

Use the natural number a to get rid of the fact that the quotient of 2008 is 44. What is the natural number and what is the remainder

2008÷44=45………… twenty-eight
So: the natural number is 45 and the remainder is 28

The quotient obtained by removing 1992 from a natural number is 46. What are the natural number and the remainder?

1992 ÷ 46 =43 ... 14
So, 1992 △ 43 = 46... 14
That is, the natural number is 43 and the remainder is 14

Absolute value of 1 / 3 minus 1 plus absolute value of 1 / 5 plus 2011 minus 1 / 2009

Absolute value of 1 / 3 minus 1 plus absolute value of 1 / 5 plus 2011 minus 1 / 2009
=1-1/3+1/3-1/5+…… +1/2009-1/2011
=1-1/2011
=2010/2011

There is a mathematical problem: A, B, C are non-zero rational numbers to find the value of a │ a │ / a │ B │ + B │ C │ / b │ C + C │ a │ / C │ a
Add a + B + C = 0, in addition, the title is a little wrong, a │ A / a │ B │ is changed to a │ A / a │ B

According to the meaning of the title: A, B, C are not zero
1. The former is positive or negative
2. A, B, C are positive and a negative. Let a, B be positive and C be negative
3. A, B and C are two negative and one positive. Let a be positive, B and C be negative, and the original formula = 1-1-1 = - 1

A + B + C = 0, the sum of A-B | a | and a-c | B |

1
method:
1. Verification method, just find a number that satisfies a + B + C = 0, such as a = 2, B = - 1, C = - 1; take a look
2. Hypothesis method, the absolute value is positively and negatively related,
Let's assume, for example, (a > 0, b > 0), (a > 0, b > 0)

b. C belongs to R +, C / (a + b) + A / (B + C) + B / (a + C) is greater than or equal to 3 / 2
Prove the formula

c/(a+b)+a/(b+c)+b/(a+c)
>=3{[c/(a+b][a/(b+c)][b/(a+c)]}^(1/3)
=3(abc)^(1/3)/[(a+b)(b+c)(c+a)]^(1/3)
>=3(abc)^(1/3)/[2(ab)^(1/2) *2(bc)^(1/2) *2(ca)^(1/2)]^(1/3)
=3(abc)^(1/3)/[2(abc)^(1/3)]
=3/2

A math problem: when a + B + C = 0, how much is a [1 / B + 1 / C] + B [1 / A + 1 / C] + C [1 / A + 1 / b]?

a+b+c=0
So a + B = - C
b+c=-a
c+a=-b
The original formula = A / B + A / C + B / A + B / C + C / A + C / b
=(a+b)\c+(b+c)\a+(c+a)\b
=-c\c+(-a)\a+(-b)\b
=-1-1-1
=-3

A mathematical problem: a + B = C, a * b = C, and a is not equal to 0, B is not equal to 0, find a, B?
I know the answer, a = b = 2, I want to know the operation process, please list, do not use iterative method, iterative method I can also find out

It should be a positive integer solution
A+B=C=A*B
AB-A-B=0
AB-A-B+1=1
(A-1)(B-1)=1
A-1=B-1=1
A=B=2

Given a = (1,2) B = (5,8) C = (2,3), what is just a * (b * c) equal to?

b*c=5*2+8*3=10+24=34
a*(b*c)=(34,68)

1. Given a = - 2, B = - 1 / 2, the value of formula - A ^ 2-AB + A / B-B ^ 2 is (). 2. Calculate (- 0.125) ^ 2008 * (- 8) ^ 2

-a^2-ab+a/b-b^2
=-4-1+4-1/4
=-3/4
（-0.125）^2008*（-8）^2
=(-1/8)^2*(-8)^2*(-1/8)^2006
=-1/(8^2006)