问题:
用换元法解方程
6(x+1)
x2+1
+
x2+1
x+1
=7,若设
x2+1
x+1
=y,则原方程可化为( )
A.y2-7y+6=0
B.y2+6y-7=0
C.6y2-7y+1=0
D.6y2+7y+1=0
用换元法解方程
+
=7,若设
=y,则原方程可化为( )
6(x+1) |
x2+1 |
x2+1 |
x+1 |
x2+1 |
x+1 |
A.y2-7y+6=0 | B.y2+6y-7=0 | C.6y2-7y+1=0 | D.6y2+7y+1=0 |
参考答案: