Question

两道初一因式分解题、、要过程，急、

1、已知ab≠0,a^2+ab-2b^2=0,那么(2a-b)/(2a+b)=
2、若2X^2+2XY+Y^2-2X+1=0,则xy=

Answer 1

1、已知ab≠0,a^2+ab-2b^2=0,
（a-b)(a+2b)=0
a=b或a=-2b
那么(2a-b)/(2a+b)=1/3或5/3
2、若2X^2+2XY+Y^2-2X+1=0
(x+y)^2+(x-1)^2=0
x+y=0 x-1=0
x=1 y=-1
,则xy=-1

Answer 2

1、∵a^2+ab-2b^2=0
∴(a+2b)(a-b)=0
∴a+2b=0或a-b=0
(2a-b)/(2a+b)=5/3或1/3
2、2X^2+2XY+Y^2-2X+1=0
x^2+2xy+y^2+x^2-2x+1=0
(x+y)^2+(x-1)^2=0
x+y=0且x-1=0
x=1,y=-1
xy=-1

Answer 3

1.
a^2+ab-2b^2=0
(a-b)(a+2b)=0
a=b或a=-2b
(2a-b)/(2a+b)=1/3或5/3
2.
2X^2+2XY+Y^2-2X+1=0
(x^2-2x+1)+(x^2+2xy+y^2)=0
(x-1)^2+(x+y)^2=0
x=1,y=-1
xy=-1