(1/x^2+3x+2)+(1/x^2+5x+6)+(1/x^2+7x+12)+(1/x^2+9x+20)

学习 时间:2026-04-07 19:46:31 阅读:4369
(1/x^2+3x+2)+(1/x^2+5x+6)+(1/x^2+7x+12)+(1/x^2+9x+20)

最佳回答

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2026-04-07 19:46:31

(1/x^2+3x+2)+(1/x^2+5x+6)+(1/x^2+7x+12)+(1/x^2+9x+20)=1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+1/(x+4)(x+5)=1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x-4)+1/(x+4)-1/(x+5)=1/(x+1)-1/(x+5)=(x+5-x-1)/(x+1)(x+5)=4/(x²+6x+5)

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  • 要减肥的睫毛膏
    回复
    2026-04-07 19:46:31

    (1/x^2+3x+2)+(1/x^2+5x+6)+(1/x^2+7x+12)+(1/x^2+9x+20)=1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+1/(x+4)(x+5)=1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x-4)+1/(x+4)-1/(x+5)=1/(x+1)-1/(x+5)=(x+5-x-1)/(x+1)(x+5)=4/(x²+6x+5)

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