So the absolute value of the set of positive integers less than 5 is less than 4, so the set of integers has enumeration method Set description method of all real numbers with absolute value less than 4 Please help me to take it,

So the absolute value of the set of positive integers less than 5 is less than 4, so the set of integers has enumeration method Set description method of all real numbers with absolute value less than 4 Please help me to take it,

Enumeration is to write out every element in a set, which is suitable for finite sets
The set of all positive integers less than 5 {1,2,3,4}
The set of all integers with absolute value less than 4 {- 3, - 2, - 1,0,1,2,3}
Description method is to extract the commonness of elements in a set and limit the scope
The set of real numbers with absolute value less than 4 {x | x | 4, X ∈ r}

Finding the value of (5a + ab-5b) / (a-ab-b) with 1 / A-1 / b = 3

Because 1 / A-1 / b = 3, B-A = 3AB (5a + ab-5b) / (a-ab-b) = (Ab-5 (B-A)) / (- ab - (B-A)) = (ab-15ab) / (- ab-3ab) = 7 / 2

If 2 (a-b) = 8, then 5b-5a =?

2（a-b)=8
The results show that: (a-b) = 4
5b-5a
=5（b-a）
=5*（-4）
=-20

A is one eighth of B. If * () is regarded as unit 1, then 5A = () B; 5B = () a

5a＝（5/8）b,5b＝40a

If 5A = 3.5b, then a: B = (): ()
Can I only fill in the simplest integer ratio? Can I fill in 3.5:5

If 5A = 3.5b, then a: B = (7): (10)
To fill in the simplest integer ratio

If the rational numbers a and B are opposite to each other, what is the value of 5a-2 + 5B?

-2

If a and B are opposite to each other, find a + 2A + +The value of 5A + 5B + 4B + B

a+2a+3a+4a+5a+5b+4b+3b+2b+b
=a+b+2a+2b+3a+3b+4a+4b+5a+5b
=(a+b)+2(a+b)+3(a+b)+4(a+b)+5(a+b)
=15(a+b)
∵ A and B are opposite to each other
∴a+b=0
The original formula is 0

Given that the absolute value of 5A + B and the square of (a + 5B + 6) are opposite, then a + B =?

This is too simple. The absolute value of 5A + B must be greater than or equal to zero, and the square of a + 5B + 6 must be greater than or equal to zero. If both numbers are positive, they cannot be opposite to each other. So we can get 5A + B = 0, 5B + A + 6 = 0, solve the system of linear equations of two variables, and get a = 0.25, B = - 1.25, so a + B = - 1

If a * B is not equal to 1, and 2A ^ 2 + 5A + 1 = 0, B ^ 2 + 5B + 2 = 0, then √ AB + (1 / √ AB) =?
AB under radical

∵2a²+5a+1=0
Obviously a ≠ 0
∴2+(5/a)+(1/a²)=0
And B & # 178; + 5B + 2 = 0
1 / A and B are two parts of the equation x & # 178; + 5x + 2 = 0
∴(1/a)+b=-5,b/a=2
∴1+ab=-5a,b=2a

If 2 / 5A = B (a, B are not equal to 0), then a: B = (): ()

2/5A=B
A:B=5:2