m+ 1/m = 1->； m^2+ 1=m->； M 2-m =- 1 Multiply both sides by m at the same time, then?

m^3-m^2=-m->； m^3=m^2-m=- 1？

m^2008=m*m^2007=m*m^(3*669)=m*(- 1)669=-m？

m^2009=(m^20 10)/m=[m^(3*670)]/m=(- 1)^670/m= 1/m？

m^2008+ 1/m^2009=-m+ 1/( 1/m)=-m+m=0？

2. Is the solution as shown in the figure?

Since the positive integer solution is considered, it is obvious that X 1, X2, X3, X4, X5, X6 >；; = 1?

So x1+x2+x3+3x4+4x5+5x6 > = 15?

That is, the minimum value of X 1+X2+X3+3X4+4X5+5X6 is 15?

X1+x2+x3+3x4+4x5+5x6 = 21The current requirement is to add 6 to the minimum value?

Why don't we set x1'= x2' = x3' = x4' = X5' = X6' =1?

So x1'+x2'+x3'+3x4'+4x5'+5x6' =15.

Let Xn=Xn'+xn(n= 1, 2, 3, 4, 5, 6), Xn >；; =0, xn is an integer.

x 1+X2+X3+3 x4+4x 5+5x 6 = x 1 '+X2 '+X3 '+3 x4 '+4x 5 '+5x 6 '+？ x 1+x2+x3+3 x4+4x 5+5x 6 = 2 1

X 1+x2+x3+3x4+4x5+5x6=6.

So now the question becomes?

x 1+x2+x3+3x4+4x5+5x6=6，？

Find the number of groups of non-negative integer solutions of xn

See the figure for other solutions (this sign is too troublesome)?

Note: The solution shown in the figure omits a set of solutions, that is, when x4=2, everything else is equal to 0, so there are 45 sets of solutions in total.