1/(x^2+x)+1/(x^2+3x+2)+1/(x^2+5x+6)+1/(x^2+7x+12)

f1asci2022-10-04 11:39:542条回答

1/(x^2+x)+1/(x^2+3x+2)+1/(x^2+5x+6)+1/(x^2+7x+12)
额.好长的题目.

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共2条回复

-白昼之月- 共回答了20个问题 | 采纳率90%: 1/(x^2+x)+1/(x^2+3x+2)+1/(x^2+5x+6)+1/(x^2+7x+12)
=1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)
=1/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)
=1/x-1/(x+4)
=4/(x^2+4x).; 1年前

表妹78 共回答了2个问题 | 采纳率: =1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)
=1/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)
=1/x-1/(x+4)
=4/(x^2+4x).; 1年前

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