Factorization: (x + y) ^ 2-2m (x + y) + m ^ 2

Factorization: (x + y) ^ 2-2m (x + y) + m ^ 2

(x+y)^2-2m(x+y)+m^2=（x+y-m）²

(1) Fill in the appropriate terms in brackets: A-B + 2C = a - () (2) calculate: 2Ab * (the square of 3a, the square of b-2ab) = ()
(1) Fill in the appropriate terms in brackets: A-B + 2C = a - () (2) calculation: 2Ab * (the square of 3a, b-2ab) = () (3) calculation: (one fourth X - two thirds x, y) * (- 12xy) = () (4) calculation: (x + 1) (x + 2) = () (5) factorization: the square of X + X-30 = () (6) calculation: (4 * 5) cube=（ ）(in scientific notation)

(1) fill in the appropriate items in brackets: A-B + 2C = a - (b-2c)
(2) Calculation: 2Ab * (the square of 3a, the square of b-2ab) = (6a ^ 3B ^ 2-4a ^ 2B ^ 3)
(3) Calculation: (quarter X - two thirds x squared y) * (- 12xy) = (- 3x ^ 2Y + 8x ^ 3Y ^ 2)
(4) Calculation: (x + 1) (x + 2) = (x ^ 2 + 3x + 2)
(5) The square of factor X + X-30 = ((X-5) (x + 6))
(6) Calculation: cube of (4 * 5) = (8 * 10 ^ 3) (expressed by scientific notation)

9A's Square X's Square - B's Square Y's Square

9A's Square X's Square - B's Square Y's Square
=The square of (3ax) the square of (by)
Square difference
=(3ax—by)(3ax+by)

Square of 9A square of (B-C)

Factor 9A & # 178; - (B-C) &# 178; = (3a + B-C) (3a-b + C)
If the brackets are removed:
9a²-(b-c)²
=9a²-b²-c²+2bc

Given that a and B are coprime numbers and 9A + 18B is the square of a single digit, how many pairs of numbers can a and B take

A. B is the coprime number,
A,B≥2
A+2B>4
9A+18B
=9 (a + 2b), the square of one digit
A + 2B = 9, a is odd
B = 2, a = 5
There is only one set of solutions

It is known that a and B are coprime numbers, and 9A + 18B is the square of one digit
Given that a and B are coprime numbers and 9A + 18B is the square of a single digit, then there are () pairs of numbers a and B can take?

In the range of positive integers, otherwise there is no finite solution
9a+18b=9(a+2b)
9 is a square number, only need a + 2B = 1,4 or 9
A + 2B = 1, only (a, b) = (1,0)
A + 2B = 4, only (a, b) = (0,2)
A + 2B = 9, there are (a, b) = (1,4), (3,3), (4,1)
So there are five pairs

Given that a and B are coprime numbers and 9A + 48b is the square of a single digit, how many pairs of numbers can a and B take?
It's 9A + 18B

9A + 18B = x ^ 2 ∵ the left side of the equation can be changed into 9 (a + 2b) ∵ the X on the right side must be a multiple of 3 and ∵ x is a single digit ∵ x can only be 3, 6, 9 when x = 3, the original formula can be changed into a + 2B = 1, rounded off when x = 6, the original formula can be changed into a + 2B = 4B = 1, a = 2B = 2, a = 0 (rounded off) when x = 9, the original formula can be changed into a + 2B = 9b = 1, a = 7b = 2, a = 5B =

A and B are a pair of Coprime numbers, AB equals 36, so what are a and B respectively

ab=36
36=18*2=4*9=12*3
Only 4 and 9 are coprime numbers
So a and B are 4 and 9
A = 4, B = 9 or a = 9, B = 4

Given that the least common multiple of two coprime numbers is 153, what are these two coprime numbers? At the same time, are 1 and 153 their coprime numbers?

1) The least common multiple of two coprime numbers is 153, and the two coprime numbers are 3,53
2) 1 and 153 are not their coprime numbers. 1 is neither a composite nor a prime number. 153 is a composite. 1 and itself are not coprime numbers

Calculation: ab-b times (A's Square - AB)

I guess what you mean is ab * (a ^ 2-AB) / (ab-b ^ 2), so it's very simple. In the two brackets, a and B are proposed respectively, and the rest are a-b. under the condition that a is not equal to B, the final result a ^ 2 (that is, the square of a) can be obtained