分类：教育/科学-学习帮助
问题：高二数学解方程:(3/x)+1/(x
解方程:
(3/x)+1/(x-1)+1/(x-2)+1/(x-3)+1/(x-4)+3/(x-5)=0
回答： (3/x)+1/(x-1)+1/(x-2)+1/(x-3)+1/(x-4)+3/(x-5)=0 
[(3/x)+3/(x-5)]+[1/(x-1))+1/(x-4)]+[1/(x-2)+3/(x-3)]=0 
[3(2x-5)/x(x-5)]+[(2x-5)/(x-1)(x-4)]+[4(2x-5)/(x-2)(x-3)]=0 
(2x-5)[3/x(x-5)+1/(x-1)(x-4)+4/(x-2)(x-3)]=0 
2x-5=0 ===>x1=5/2 
或3/x(x-5)+1/(x-1)(x-4)+1/(x-2)(x-3) =0
==>
3/(x²-5x) +1/(x²-5x+4) +4/(x²-5x+6) =0
换元,x²-5x+4=T
==>3/(T-3) +1/T +4/(T+2) =0
==>2T²-3T-2=0 ==>T=2,T=-1/2
T=2 ===>x =(5 ±√17)/2
T=-1/2 ==>x=(5 ±√7)/2
显然上述5个解都不是增根
==>方程的解
x1=5/2 ;
x2,3 =(5 ±√17)/2
x4,5=(5 ±√7)/2