首页
(2004•扬州)用换元法解方程(x+)2-(2x+)=3,则原方程可化为A.y2+2y-...
试题详情
(2004•扬州)用换元法解方程(x+
![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAAsAAAAjCAYAAABCU/B9AAAAAXNSR0IArs4c6QAAAARnQU1BAACx
jwv8YQUAAAAJcEhZcwAADsMAAA7DAcdvqGQAAACuSURBVDhP7ZSLDcMwCESZjJ1Yh2UYxaNQzglO
3To/palUqZasyNIF8PEw+YlFJ7R+UVzUmdi1vOfsI5s4EcU+Iq7BzOUvbrYedaM2BD5PW6xvzMV2
b5B1Y+S8zN4X1d1YxhfH6nM+59yx+jSjACnYaOcl1WxdcWUMKYQS8vFqPhflIG1d2DcFeA5SP+do
kS0irz0u+UMVowSAbjIDH5d+Bb+WAYA436p0ZqT8UZAeDDeKw2COEx0AAAAASUVORK5CYII=
)
)
2-(2x+
![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAAsAAAAjCAYAAABCU/B9AAAAAXNSR0IArs4c6QAAAARnQU1BAACx
jwv8YQUAAAAJcEhZcwAADsMAAA7DAcdvqGQAAADVSURBVDhPzZTrDYMwDIQ9mXfKOlnGo2QUEyvY
YMcpQpSqSPygPS7nxwfwjQtuaPmBmAoDwLixcgvHHs6tMhba/yYu8oI9j59N3IicU6s4ua8zS6SV
c+wKFeCgXXTD5T9skhiNKxbWUs8nTmIqyDX2bH/DiaUDLmcv8vxsYinIBqKDAR/nwbgvtupF57mo
rNAR4MUYP8Lqu31WSA3Qnb81sLLDspoizHfZAyuAfhD6oQhKydEpKdSdEdaUmLMSYkQHQtS97xEw
KnTamfgNyBi8Gv2fbN0GHxCBOu78AsgAAAAASUVORK5CYII=
)
)=3,则原方程可化为( )
A.y
2+2y-3=0
B.y
2-2y+3=0
C.y
2-2y-3=0
D.y
2+2y+3=0
相关试题
-
(2004•扬州)用换元法解方程(x+
)2-(2x+
)=3,则原方程可化为( )
A.y2+2y-3=0
B.y2-2y+3=0
C.y2-2y-3=0
D.y2+2y+3=0
-
(2004•扬州)用换元法解方程(x+
)2-(2x+
)=3,则原方程可化为( )
A.y2+2y-3=0
B.y2-2y+3=0
C.y2-2y-3=0
D.y2+2y+3=0
-
(2004•盐城)解分式方程
时,可设
=y,则原方程可化为整式方程是( )
A.y2+2y+1=0
B.y2+2y-1=0
C.y2-2y+1=0
D.y2-2y-1=0
-
(2004•盐城)解分式方程
时,可设
=y,则原方程可化为整式方程是( )
A.y2+2y+1=0
B.y2+2y-1=0
C.y2-2y+1=0
D.y2-2y-1=0
-
解方程
时,若
=y,则原方程可化为( )
A.y2-2y-1=0
B.y2-2y-3=0
C.y2-2y+1=0
D.y2+2y-3=0
-
解分式方程
时,可设
=y,则原方程可化为整式方程是( )
A.y2+2y+1=0
B.y2+2y-1=0
C.y2-2y+1=0
D.y2-2y-1=0
-
(2006•咸宁)解方程:x2+
-1=0时,若设x+
=y,则原方程可化为( )
A.y2-2y-1=0
B.y2-2y-3=0
C.y2-2y+1=0
D.y2+2y-3=0
-
(2006•咸宁)解方程:x2+
-1=0时,若设x+
=y,则原方程可化为( )
A.y2-2y-1=0
B.y2-2y-3=0
C.y2-2y+1=0
D.y2+2y-3=0
-
如用换元法解方程
,那么原方程可化为( )
A.y2-3y+2=0
B.y2+3y-2=0
C.y2-2y+3=0
D.y2+2y-3=0
-
如用换元法解方程
,那么原方程可化为( )
A.y2-3y+2=0
B.y2+3y-2=0
C.y2-2y+3=0
D.y2+2y-3=0