（1997•甘肃）用换元法解分式方程（xx−1）+2（x−1x）+3=0，应该设y=xx−1或x−1xxx−1或x−1x．

Question

（1997•甘肃）用换元法解分式方程（(x)/(x−1)）+2（(x−1)/(x)）+3=0，应该设y=

(x)/(x−1)或(x−1)/(x)

(x)/(x−1)或(x−1)/(x)

．

Answer 1

用换元法解分式方程（(x)/(x−1)）+2（(x−1)/(x)）+3=0，应该设y=(x)/(x−1)或(x−1)/(x)．
故答案为：(x)/(x−1)或(x−1)/(x)