问题描述: 用部分分式的方法,解方程1除于(x^2+3x+2)+1除于(x^2+5x+6)+1除于(x^2+7x+12)=1除于3x 1个回答 分类:数学 2014-10-21 问题解答: 我来补答 ∵1/(x^2+3x+2)+1/(x^2+5x+6)+1/(x^2+7x+12)=1/(3x),∴1/[(x+1)(x+2)]+1/[(x+2)(x+3)]+1/[(x+3)(x+4)]=1/(3x),∴1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)=1/(3x),∴1/(x+1)-1/(x+4)=1/(3x),∴[(x+4)-(x+1)]/[(x+1)(x+4)]=1/(3x),∴(x+1)(x+4)=9x,∴x^2+5x+4=9x,∴x^2-4x+4=0,∴(x-2)^2=0,∴x=2.经检验,x=2是原方程的解. 展开全文阅读