x-2的绝对值+y-4的绝对值=0则xy/1+(x+2)(y+2)/1+(x+4)(y+4)/1+...+(x+1994)(y+1994)/1=

绝对值项恒非负,两绝对值项之和=0,两绝对值项分别=0
x-2=0 x=2
y-4=0 y=4
y=x+2
1/(xy)+1/[(x+2)(y+2)]+1/[(x+4)(y+4)]+...+1/[(x+1994)(y+1994)]
=1/[x(x+2)]+1/[(x+2)(x+4)]+1/[(x+4)(x+6)]+...+1/[(x+1994)(x+1996)]
=(1/2)[1/x-1/(x+2)+1/(x+2)-1/(x+4)+1/(x+4)-1/(x+6)+...+1/(x+1994)-1/(x+1996)]
=(1/2)[1/x -1/(x+1996)]
=(1/2)(1/2 -1/1998)
=499/1998