若|x-1|+|y-3|=0,求1/xy+1/(x+2)(y+2)+1/(x+4)(y+4)+1/(x+6)(y+6）.

ccx19822022-10-04 11:39:541条回答

若|x-1|+|y-3|=0,求1/xy+1/(x+2)(y+2)+1/(x+4)(y+4)+1/(x+6)(y+6）.+1/(x+2010)(y+2010)的值!我做了,就是最后两步错了,不知道怎么错了,我把X=1,Y=3带入最后得到1/1*3+1/3*5+1/5*7+...+1/2011*2013=1-1/2013=2012/2013,老师给我扣了三分,

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gycgyc 共回答了19个问题 | 采纳率94.7%: 绝对值项恒非负,两绝对值项之和=0,两绝对值项分别=0
x-1=0 x=1
y-3=0 y=3
y=x+2
1/(xy)+1/[(x+2)(y+2)]+1/[(x+4)(y+4)]+...+1/[(x+2010)(y+2010)]
=1/[x(x+2)]+1/[(x+2)(x+4)]+1/[(x+4)(x+6)]+...+1/[(x+2010)(x+2012)]
=(1/2)[1/x -1/(x+2)+1/(x+2)-1/(x+4)+1/(x+4)-1/(x+6)+...+1/(x+2010)-1/(x+2012)]
=(1/2)[1/x -1/(x+2012)]
=(1/2)[1/1 -1/(1+2012)]
=(1/2)(1-1/2013)
=1006/2013

老师扣你分的原因是你算错了.1/[n(n+2)]=(1/2)[1/n -1/(n+2)]而不是1/[n(n+2)]=1/n -1/(n+2); 1年前

当丨x-1丨+丨y-3丨=0时,探究并计算:1/xy+1/(x+2)(y+2)+1/(x+4)( 当丨x-1丨+丨y-3丨=0时,探究并计算:1/xy+1/(x+2)(y+2)+1/(x+4)( +4)+1/(x+6) sea_rover1年前1: 9271891 共回答了16个问题 | 采纳率75%

∵丨x-1丨+丨y-3丨=0
∴x-1=0 y-3=0
∴x=1 y=3
∴原式=1/(1×3)+1/(3×5)+1/(5×7)+1/(7×9)×
=1/2(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9)
=1/2(1-1/9)
=1/2×8/9
=4/9

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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/( 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) ygwmt1年前1: rainofsunday 共回答了17个问题 | 采纳率82.4%

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

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若|x-1|+|y-3|=0,求1/xy+1/(x+2)(y+2)+1/(x+4)(y+4)+1/(x+6)(y+6）.

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