赞 whitebeard 幼苗
共回答了12个问题采纳率：91.7% 举报
(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)
=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)
1年前
3
共回答了15个问题 举报
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/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
1
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