If 1-3m-3n = 2, then n + M=

If 1-3m-3n = 2, then n + M=

1-3m-3n=2
3m＋3n=1-2=-1
m＋n=-1/3

If 3M + 2 = 3N + 1, the relationship between M and N is

3m+2=3n+1
3m-3n=1-2
3(m-n)=-1
m-n=-1/3
M

When m > n is known, the size of 2-3m and 3-3n can be judged

Because m > N, 3M > 3N, so - 3M

If 3m-2n-1 = 3n-2m, then the relationship between M and N is m＞n m＜n M = n cannot be determined

3m-2n-1=3n-2m
5m-5n=1
5(m-n)=1
m-n=0.2
M>n

The rational number m and N are opposite to each other, the product of X and Y is 1, and the absolute value of Z is 7. Find the value of 3M + 3N + 5 + 5xy + Z

m+N=0 xy=1 z=±7 3m+3N+5+5xy+Z=5+5+7=17 3m+3N+5+5xy+Z=5+5-7=3

The rational number Mn is opposite to each other, XY is reciprocal to each other, and the absolute value of Z is equal to 7

Mn is opposite to each other = > m + n = 0
XY reciprocal = > xy = 1
The absolute value of Z is equal to 7 = > z = 7, - 7
3m+3n+5xy+z
=3(m+n)+5xy+z
=-2 or 12

How does the root sign (n? - 3n) (n? - 3N + 2) + 1 be transformed into the radical sign (n? - 3n) 2 + 2 (n? - 3n) + 1?

Let a = n? - 3N
Then the original formula = √ [a (a + 2) + 1]
=√(a²+2a+1)
=√[(n²-3n)²+(n²-3n)+1]

(m-n)(3m-n)²+(m+3n)²(n-m) Tonight! Help me!

(m-n)(3m-n)²+(m+3n)²(n-m)
=(m-n)(3m-n)²+[-(m+3n)²(m-n)]
=(m-n)(3m-n)²-(m+3n)²(m-n)
=(m-n)[(3m-n)²-(m+3n)²]
=(m-n)[(3m-n)+(m+3n)][(3m-n)-(m+3n)]
=(m-n)(4m+2n)(2m-4n)
=4(m-n)(2m+n)(m-2n)

Given 4m + n = 9,2m-3n = 1, find the value of (M + 2n) 2 - (3m-n) 2

Because 4m + n = 9, 2m-3n = 1
So by subtracting the two formulas, we get: 2m + 4N = 8, M + 2n = 4,
By adding the two formulas, 6m-2n = 10, 3m-n = 5
So (M + 2n) 2 - (3m-n) 2 = 4? - 5? = (4 + 5) * (4-5) = - 9

It is known that | m-2 | + (n + 1) 5) 2 = 0, find the value of M - (M 2n + 3m-4n) + 3 (2nm2-3n)

∵|m-2|+（n+1
5）2=0，
∴m=2，n=−1
5，
The original formula = m-m2n-3m + 4N + 6nm2-9n
=5m2n-2m-5n    
When − M = 1
At 5:00,
Original formula = 5 × 22 × (- 1)
5）-2×2-5×（-1
5）  
=-4-4+1
=-7．