（m－a）²－（n＋b）²

（m－a）²－（n＋b）²

Original formula = [(M-A) + (n + b)] * [(M-A) - (n + b)]
=(m+n-a+b) * (m-n-a-b)

What is M & # 178; + m-2
Factorization I know the answer is (M - 1) (M + 2), but I forgot how to find it. I remember it was a big square made up of several rectangles

m²+m-2
=m²-2m+1+3m-3
=(m-1)²+3(m-1)
=(m-1)(m-1+3)
=(m-1) (M + 2) I hope I can help you

M & # 178; = 1, is m equal to ± √ 1

It's ± √ 1, which is ± 1

(x-m-2) &# (178; how much is it to remove brackets

X²+m²+4-2xm-4x+4m

Given the set M = {x | x = m + 1 / 6, m ∈ Z}, P = {x | 2 / P + 1 / 6, P ∈ Z}, then the relation between M and P is?

Because M = {x | x = m + 1 / 6, m ∈ Z}, P = {x | 2 / P + 1 / 6, P ∈ Z}, so m = {x | x = 6 / 6m + 1, m ∈ Z}, P = {x | 6 / 3P + 1, P ∈ Z}, and because m p ∈ Z} | 6 m | is greater than | 3P | so the probability that x ∈ Z 6m + 1 is an integer is less than 3P + 1, and 3P + 1 has at least one integer result, 6m + 1 cannot satisfy

Given the set M = {x | x-4 | + | X-1 | < 5}, n = {x | (x-a) (X-6) < 0}, and m ∩ n = (2, b), then a + B=______ ．

∫| x-4 | + | X-1 | < 5, we can see from the geometric meaning of absolute value that the sum of distances from 1 and 4 on the number axis is less than 5, ∫ 4-1 = 3, | 5-1 | + | 5-4 | = 5, | 0-1 | + | 0-4 | = 5, ∫ m = {x | 0 ＜ x ＜ 5}, n = {x | (x-a) (X-6) ＜ 0}, and m ∩ n = (2, b), ∩ a = 2, B = 5

Given that M = {A / a belongs to Z, and 6 / 5-a belongs to n}, then why is m 5-a = 1,2,3,6, so a = 4,3,2, - 1

6 / (5-a) is a natural number
So 5-a is a divisor of 6
So 5-a = 1,2,3,6
So a = 4,3,2, - 1

We know that two sets are m = {x | 1

（0,1】

Given the set M = {x | (x + 2) (x-1) ＜ 0}, n = {x | x + 1 ＜ 0}, then m ∩ n = ()
A. （-1，1）B. （-2，1）C. （-2，-1）D. （1，2）

∵set M = {x | (x + 2) (x-1) ＜ 0}, ∩ M = {x | - 2 ＜ x ＜ 1}, ∵ n = {x | x + 1 ＜ 0}, ∩ n = {x | - 2 ＜ x ＜ - 1}, so select C

Let m = (Y / y = 12 / 12 of X-12, X belongs to Z, y belongs to n), n = (x / y = 12 / 2 of X-2, X belongs to Z, y belongs to n), try to find m intersection N and M union n respectively

It is known that the number in M must be divisible by 12
M={1,2,3,4,6,12}
And n-2 must be divisible by 12
N-2={1,2,3,4,6,12}
So n = {3,4,5,6,8,14} so
M intersection n = {3,4,6}
M and N = {1,2,3,4,5,6,8,12,14}