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)分之1=1](/uploads/image/z/3805923-3-3.jpg?t=%E8%AE%A1%E7%AE%97n%28n%2B1%29%E5%88%86%E4%B9%8B1%2B%28n%2B1%29%28n%2B2%29%E5%88%86%E4%B9%8B1%2B%28n%2B2%29%28n%2B3%29%E5%88%86%E4%B9%8B1%2B%28n%2B2007%29%28n%2B2008%29%E5%88%86%E4%B9%8B1%E8%A7%A3%E5%88%86%E5%BC%8F%E6%96%B9%E7%A8%8B%EF%BC%9A%EF%BC%88x-2%EF%BC%89%E5%88%86%E4%B9%8B1%2B%5B%EF%BC%88x-2%EF%BC%89%EF%BC%88x-3%EF%BC%89%5D%E5%88%86%E4%B9%8B1%2B%EF%BC%88x-3%EF%BC%89%EF%BC%88x-4%29%E5%88%86%E4%B9%8B1%3D1)
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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