计算n(n+1)分之1+(n+1)(n+2)分之1+(n+2)(n+3)分之1+(n+2007)(n+2008)分之1解分式方程:(x-2)分之1+[(x-2)(x-3)]分之1+(x-3)(x-4)分之1=1

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计算n(n+1)分之1+(n+1)(n+2)分之1+(n+2)(n+3)分之1+(n+2007)(n+2008)分之1解分式方程:(x-2)分之1+[(x-2)(x-3)]分之1+(x-3)(x-4)

计算n(n+1)分之1+(n+1)(n+2)分之1+(n+2)(n+3)分之1+(n+2007)(n+2008)分之1解分式方程:(x-2)分之1+[(x-2)(x-3)]分之1+(x-3)(x-4)分之1=1
计算n(n+1)分之1+(n+1)(n+2)分之1+(n+2)(n+3)分之1+(n+2007)(n+2008)分之1
解分式方程:(x-2)分之1+[(x-2)(x-3)]分之1+(x-3)(x-4)分之1=1

计算n(n+1)分之1+(n+1)(n+2)分之1+(n+2)(n+3)分之1+(n+2007)(n+2008)分之1解分式方程:(x-2)分之1+[(x-2)(x-3)]分之1+(x-3)(x-4)分之1=1
(1) n(n+1)分之1+(n+1)(n+2)分之1+(n+2)(n+3)分之1+(n+2007)(n+2008)分之1
=n分之1-(n+1)分之1+(n+1)分之1-(n+2)分之1+.+(n+2007)分之1-(n+2008)分之1
=n分之1-(n+2008)分之1
=n(n+2008)分之(n+2007)
(2)(x-2)分之1+[(x-2)(x-3)]分之1+(x-3)(x-4)分之1=1
即(x-4)分之1-(x+3)分之1+(x-3)分之1-(x-2)分之1+(x-2)分之1=1
所以(x-4)分之1=1
则 x-4=1,解得 x=5

[(x-2)(x-3)]分之1 可换为1/(x-3)减去1/(x-2)
同理(x-3)(x-4)分之1可换为1/(x-4)减去1/(x-3)
综上带入的原式化为1/(x-2)+1/(x-3)-1/(x-2)+1/(x-4)-1/(x-3)
得1/(x-4)=1 就能算出结果了!
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[(x-2)(x-3)]分之1 可换为1/(x-3)减去1/(x-2)
同理(x-3)(x-4)分之1可换为1/(x-4)减去1/(x-3)
综上带入的原式化为1/(x-2)+1/(x-3)-1/(x-2)+1/(x-4)-1/(x-3)
得1/(x-4)=1 就能算出结果了!
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