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解1\x^2+3x+2+1\x^2+5x+6+1\x^2+7x+12+1/x^2+9x+20
人气:411 ℃ 时间:2019-11-06 09:02:29
解答
原式=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)
=4/(x²+6x+5)
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