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(2002•宜昌)用换元发解方程-6=0,令y=代入原方程后,变形正确的为A.y2+5=0...
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(2002•宜昌)用换元发解方程
![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAACkAAAAjCAYAAAAJ+yOQAAAAAXNSR0IArs4c6QAAAARnQU1BAACx
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)
-6
![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAJsAAAAjCAYAAACD+HiUAAAAAXNSR0IArs4c6QAAAARnQU1BAACx
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)
=0,令y=
![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABcAAAAjCAYAAAB/wZEbAAAAAXNSR0IArs4c6QAAAARnQU1BAACx
jwv8YQUAAAAJcEhZcwAADsMAAA7DAcdvqGQAAADmSURBVEhL7ZXhDYQwCIWZjJ26TpfpKB2FA0y9
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P584LjySOF14NHF+wsOJow30Hn6f+xxvos50e+CmOS+I/a3k0ZsJOQAAAABJRU5ErkJggg==
)
代入原方程后,变形正确的为( )
A.y
2+5=0
B.y
2-6y=0
C.(y+1)(y+5)=0
D.(y-1)(y-5)=0
相关试题
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(2002•宜昌)用换元发解方程
-6
=0,令y=
代入原方程后,变形正确的为( )
A.y2+5=0
B.y2-6y=0
C.(y+1)(y+5)=0
D.(y-1)(y-5)=0
-
(2002•宜昌)用换元发解方程
-6
=0,令y=
代入原方程后,变形正确的为( )
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用换元法解方程
,令
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(2002•盐城)解方程
,设y=x+
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A.y2-3y=0
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(2002•盐城)解方程
,设y=x+
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(2002•大连)解方程
,设y=
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A.y2-5y+2=0
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,设y=
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A.y2-5y+2=0
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-
(2002•大连)解方程
,设y=
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用换元法解方程
,若设
=y,则原方程可化为( )
A.y2-7y+6=0
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(2003•郴州)解方程
时,令y=x2+2x,原方程可化为( )
A.y2-5y-6=0
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