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=

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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=

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=
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=

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

|x-2|+|y-4|=0
因为|x-2|≥0,|y-4|≥0,所以当它们之和为零时,只能同时为零
则,x=2,y=4
原式1/(2*4)+1/(4*6)+1/(6*8)+……+1/(1996*1998)
=(1/2)*[(1/2)-(1/4)]+(1/2)*[(1/4)-(1/6)]+……+(1/2)*[(1/1996)-(1/1998)]
=(1/2)*...

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|x-2|+|y-4|=0
因为|x-2|≥0,|y-4|≥0,所以当它们之和为零时,只能同时为零
则,x=2,y=4
原式1/(2*4)+1/(4*6)+1/(6*8)+……+1/(1996*1998)
=(1/2)*[(1/2)-(1/4)]+(1/2)*[(1/4)-(1/6)]+……+(1/2)*[(1/1996)-(1/1998)]
=(1/2)*[(1/2)-(1/4)+(1/4)-(1/6)+……+(1/1996)-(1/1998)]
=(1/2)*[(1/2)-(1998)]
=(1/2)*(998/1998)
=499/1998

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