问题描述: 解方程1\x^+3x+2 + 1\x^2+5x+6 + 1\x^2+7x+12 + 1\x^2+x=4/21的解是 1个回答 分类:数学 2014-11-16 问题解答: 我来补答 1/(x^2+3x+2)+1/(x^2+5x+6)+1/(x^2+7x+12)+1/(x^2+x)=4/211/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+1/x(x+1)=4/211/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)=4/211/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)=4/211/x-1/(x+4)=4/21(x+4-x)/x(x+4)=4/214/x(x+4)=4/211/x(x+4)=1/21x(x+4)=21x^2+4x-21=0(x+7)(x-3)=0x=-7或x=3经检验x=-7或x=3是方程的解 再问: 方程1/x-1+2/x-2=1的解的个数是 展开全文阅读