用换元法求3⼀x+1⼀(x-1)+4⼀(x-2)+4⼀(x-3)+1⼀(x-4)+3⼀(x-5)=0

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^2-5x+4)+4/(x^2-5x+6)]=0

所以2x-5可以等于0 所以x=5/2

由[3/(x^2-5x)+1/(x^2-5x+4)+4/(x^2-5x+6)]=0

得：3(x^2-5x+4)(x^2-5x+6)+(x^2-5x)(x^2-5x+6)+4(x^2-5x)(x^2-5x+6)=0 (实际上就是把式子合并后，提出分子）

然后继续就行了

继续下去的话会得到以x^2-5x为未知值的方程，解出来就是答案了

答案一共5个 5/2 正负根号下17+5/2 正负根号下7+5/2

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
3(2x-5)/(x^2-5x)+(2x-5)/(x^2-5x+4)+4(2x-5)/(x^2-5x+6)=0
(2x-5)[3/(x^2-5x)+1/(x^2-5x+4)+4/(x^2-5x+6)]=0
2x-5可以等于0 所以x=5/2
由[3/(x^2-5x)+1/(x^2-5x+4)+4/(x^2-5x+6)]=0
设x^2-5x=y
3/y+1/(y+4)+4/(y+6)=0
3(y+4)(y+6)+y(y+6)+4y(y+4)=0
8y^2+52y+72=0
2y^2+13y+18=0
(2y+9)(y+2)=0
y=-9/2,或y=-2
x^2-5x==-9/2或x^2-5x=-2
答案共5个 5/2 正负根号下17+5/2 正负根号下7+5/2

3/x+1/(x-1)+4/(x-2)+4/(x-3)+1/(x-4)+3/(x-5)=0
3(1/x+1/(x-5))+(1/(x-1)+1/(x-4))+4(1/(x-2)+1/(x-3))=0
3(2x-5)/(x^2-5x)+(2x-5)/(x^2-5x+4)+4(2x-5)/(x^2-5x+6)=0
(2x-5)(3/(x^2-5x)+1/(x^2-5x+4)+4/(x^2-5x+6))=0
2x-5=0,x1=5/2

设x^2-5x=y
3/y+1/(y+4)+4/(y+6)=0
3(y+4)(y+6)+y(y+6)+4y(y+4)=0
8y^2+52y+72=0
2y^2+13y+18=0
(2y+9)(y+2)=0
y1=-2,y2=-9/2

y=-2时，x^2-5x=-2
x^2-5x+2=0
x=(5±√17)/2

y=-9/2时，x^2-5x=-9/2
2x^2-10x+9=0
x=(5±√7)/2