用换元法解方程x2−1x−xx2−1+2=0,如果设y=x2−1x,那么原方程可化为(  ) A.y2-y+2=0 B.y2+y-2=0 C.y2-2y+1=0 D.y2+2y-1=0

问题描述:

用换元法解方程

x2−1
x
x
x2−1
+2=0,如果设y=
x2−1
x
,那么原方程可化为(  )
A. y2-y+2=0
B. y2+y-2=0
C. y2-2y+1=0
D. y2+2y-1=0

y=

x2−1
x

则方程
x2−1
x
x
x2−1
+2=0变为y−
1
y
+2
=0,
整理得y2+2y-1=0,
故选D.