用换元法解方程x2−3x+3x2−3x=4时,设y=x2-3x,则原方程可化为(  ) A.y+3y−4=0 B.y−3y+4=0 C.y+13y−4=0 D.y+13y+4=0

问题描述:

用换元法解方程x2−3x+

3
x2−3x
=4时,设y=x2-3x,则原方程可化为(  )
A. y+
3
y
−4=0

B. y−
3
y
+4=0

C. y+
1
3y
−4=0

D. y+
1
3y
+4=0

设x2-3x=y,
则原方程可化为:
y+

3
y
=4.
即:y+
3
y
−4=0

故选A.