Knowing that there is a rational number m, n satisfies n 3 2 m m and 2m n 1 2, then m 2 3n

1. When m is greater than 0, (n+3) 2+|m|=m becomes (n+3) 2-m=m, i.e., n+3) 2=0

So n=-3, and then |2m-n+1|When 2m-n in =2 + 1 = -2, m = -3 (rounded).

So n=-3, and then |2m-n+1|When 2m-n+1=2 in =2, m=-1 (rounded).

2. When m is less than or equal to 0, (n+3) 2+|m|=m becomes (n+3) 2+m=m, i.e., (n+3) 2=2m

Then (n+3) 2 is less than 0 and there is no solution.

3. When m=0, then |2m-n+1|When 2m-n+1=-2 in =2, n=3, i.e., n+3) 2 is not 0, rounded off.

When m=0, then |2m-n+1|=2 in 2m-n+1=2, n=-1, i.e., n+3) 2 is not 0, rounded.

Based on the above, there is no solution to this problem. That is, the value of m 2+3n cannot be found.

The rational number m, n satisfies (n+3) 2+|m|=m, the right side of the equation (n+3) 2>=0, |m|>=0, so m>=0, so the equation can be reduced to (n+3) 2=0, and the solution is n=-3, because |2m-n+1|=2, substitute n=-3 into it, and get |2m+4|=2, and because m>=0, m=?

Looks like there's something wrong with the title.

Since the rational number m, n satisfies (n+3) 2+|m|=m, then m>=0, then |m|=m，n=-3

Again|2m-n+1|=2,|2m-3+1|=2, then m=0 or 2, when m=0, m2+3n=-6

When m=2, m2+3n=-2

Since the rational number m, n satisfies (n+3) 2+|m|=m, then m>=0, then |m|=m,n=-3

Again|2m-n+1|=2,|2m-3+1|=2, then m=0 or 2

Then when m=0, m2+3n=-6

When m=2, m2+3n=-2,10, the rational number m,n satisfies (n+3) 2+|m|=m, the right side of the equation (n+3) 2>=0, |m|>=0, so m>=0, so the equation can be reduced to (n+3) 2=0, and the solution is n=-3, because the code is |2m-n+1|=2, substitute n=-3 into it, and get |2m+4|=2, and because m>=0, m=?

It seems that there is a problem with the problem, the problem is fine, m is indeed buried in 0, but where to get the equation (n+3) 2=0? It should be (n+3) 2 0, which can be in the junior range, which is constant. 2,1, when m is greater than 0, (n+3) 2+|m|=m becomes (n+3) 2-m=m, i.e., n+3) 2=0

So n=-3, and then |2m-n+1|When 2m-n in =2 + 1 = -2, m = -3 (rounded).

So n=-3, and then |2m-n+1|When 2m-n+1=2 in =2, m=-1 (rounded).

2. When m is less than or equal to 0, (n+3) 2+|m|=m becomes (n+3) 2+m=m, i.e., (n+3) 2=2m

then (n+..0,

From the bridge grip of the question, it can be obtained: (m-2n) * 5 + (2m + 3n + 7) = 0

Since 5 is an irrational number, then there must be:

m-2n=0；② 2m+3n+7=0.

Therefore, m=2n, 2*2n+3n+7=0, you can get:

m=-2,n=-1.

Then the balance of Senqing is calculated as: m+n=-3

m n satisfies 2+|n^2-4|=0

Then Lingyuan Yuan can only collapse by foot.

m+n 4=0 and n 2-4=0

Splitting n=2, m=-1 2

or n=-2, m=1 2

There is always mn=-1

Then m 3n 3=(mn) 3=(-1) 3=-1

The root is not covered with repentance, according to the meaning of the stuffy shed, 2m-1=0, n+2=0, and the solution is m=<>

n=-2, so, mn=<>

So the answer is: -1 withering.

m=-7/2,n=-7/4

The key to this problem is that the addition of integers is equal to 0, and the addition of the root number 5 is equal to 0, so 2m+7=0 m=-7 2

7 2) root number 5 = 2n root number 5 so n = -7 4

From the question formula, it can be obtained: (m-2n)* 5+(2m+3n+7)=0.

Since 5 is an irrational number, then there must be:

m-2n=0；② 2m+3n+7=0.

Therefore, m=2n, 2*2n+3n+7=0, you can get:

m=-2，n=-1.

Then it is obtained: m+n=-3