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

学习 时间:2026-08-15 10:16:10 阅读:612
(1/x^2+3x+2)+(1/x^2+5x+6)+(1/x^2+7x+12)+(1/x^2+9x+20)

最佳回答

激昂的发带

大胆的微笑

2026-08-15 10:16:10

(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)

最新回答共有2条回答

  • 复杂的书本
    回复
    2026-08-15 10:16:10

    (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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