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已知x-2的绝对值+x-4的绝对值=0.求1/xy+1/（x+2)*(y+2)+1/(x+4)*(y+4)+.+x1/(x+2010)*(y+2010)
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已知x-2的绝对值+x-4的绝对值=0.求1/xy+1/（x+2)*(y+2)+1/(x+4)*(y+4)+.+x1/(x+2010)*(y+2010)
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答案和解析
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=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
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