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+
=4时,设y=x2-3x,则原方程可化为( )3
x2−3x
A. y+
−4=03 y
B. y−
+4=03 y
C. y+
−4=01 3y
D. y+
+4=0 1 3y
答
设x2-3x=y,
则原方程可化为:
y+
=4.3 y
即:y+
−4=0.3 y
故选A.