用换元法求3/x+1/(x-1)+4/(x-2)+4/(x-3)+1/(x-4)+3/(x-5)=0

题目

用换元法求3/x+1/(x-1)+4/(x-2)+4/(x-3)+1/(x-4)+3/(x-5)=0
需要好好想想.有答案5/2,(5±√17)/2,(5±√7)/2.

答案

这道题要求计算能力很强 3/x+1/(x-1)+4/(x-2)+4/(x-3)+1/(x-4)+3/(x-5)=0 [3/x+3/(x-5)]+[1/(x-1)+1/(x-4)]+[4/(x-2)+4/(x-3)]=0 (6x-15)/(x^2-5x)+(2x-5)/(x^2-5x+4)+(8x-20)/(x^2-5x+6)=0 (2x-5)[3/(x^2-5x)+1/(x^...