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