已知x+y=1,xy=-2分之1,利用因式分解求:x(x+y)(x-y)-x(x+y)²的值

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已知x+y=1,xy=-2分之1,利用因式分解求:x(x+y)(x-y)-x(x+y)²的值

已知x+y=1,xy=-2分之1,利用因式分解求:x(x+y)(x-y)-x(x+y)²的值
已知x+y=1,xy=-2分之1,利用因式分解求:x(x+y)(x-y)-x(x+y)²的值

已知x+y=1,xy=-2分之1,利用因式分解求:x(x+y)(x-y)-x(x+y)²的值
已知:x y=1,xy=-2分之1,利用因式分解求x(x y)(x-y)-x(x y)^2的值 x(x y)(x-y)-x(x y)^2 =x(x y)(x-y-x-y) =-2xy(

原式=x(x+y)[(x-y)-(x+y)]=x(x+y)(x-y-x-y)=x(x+y)(-2y)=-2xy(x+y)
把x+y=1,xy=-½代入上式中,得
-2×(-½)×1=1×1=1

原式=x(x+y)(x-y-x-y)=-2xy(x+y)=1*1=1

原式=x(x+y)[(x-y)-(x+y)]=x(x+y)(x-y-x-y)=x(x+y)(-2y)=-2xy(x+y)
把x+y=1,xy=-½代入上式中,得
-2×(-½)×1=1×1=1