化简 (1/x^2+3x+2)+(1/x^2+5x+6)+(1/x^2+4x+3) （化积为差，裂项相消）
展开全部
(1/x^2+3x+2)+(1/x^2+5x+6)+(1/x^2+4x+3)
=1/[(x+1)(x+2)]+1/(x+2)(x+3)]+1/[(x+1)(x+3)]
=1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/2[1/(x+1)-1/(x+3)]
=[1/(x+1)-1/(x+3)]×3/2
=3/(x²+4x+3)
=1/[(x+1)(x+2)]+1/(x+2)(x+3)]+1/[(x+1)(x+3)]
=1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/2[1/(x+1)-1/(x+3)]
=[1/(x+1)-1/(x+3)]×3/2
=3/(x²+4x+3)
本回答由提问者推荐
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
