(x^-5)/(x-1)+(10x-10)/(x^-5)=7
令k=(x^-5)/(x-1)则原式转化为
k+10/k=7
k^+10=7k
k^-7k+10=0
k=2或5
由k=(x^-5)/(x-1)=2解得x=-1或3
由k=(x^-5)/(x-1)=5解得x=0或5
经检验都符合条件所以x=-1或3或0或5