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

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

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

已知:x+y=1,xy=-1/2,利用因式分解求:x(x+y)(x-y)-x(x+y)^2
x(x+y)(x-y)-x(x+y)^2的公因式为x(x+y),提公因式得:
原式=x(x+y)[(x-y)-(x+y)]=x(x+y)(x-y-x-y)=-2xy(x+y),
代入已知条件x+y=1,xy=-1/2得
原式=(-2)*(-1/2)=1

x(x+y)(x-y)-x(x+y)^2的公因式为x(x+y),提公因式得:
原式=x(x+y)[(x-y)-(x+y)]=x(x+y)(x-y-x-y)=-2xy(x+y),
代入已知条件x+y=1,xy=-1/2得
原式=(-2)*(-1/2)=1

x(x+y)(x-y)-x(x+y)^2
=x(x+y)[(x-y)-(x+y)]
=x(x+y)(-2y)
=-2xy(x+y)
=-2*(-1/2)*1
=1

x(x+y)(x-y)-x(x+y)^2=x(x+y)[(x-y)-(x+y)]=x(x+y)(-2y)=-2xy(x+y)
代入得:-2*1*(-1/2)=1

原式=x(x+y)(x-y)-x(x+y)(x+y)
=x(x+y)[(x-y)-(x+y)]
=x(x+y)(x-y-x-y)=
-2xy(x+y),
代入已知条件x+y=1,xy=-1/2得
原式=(-2)*(-1/2)=1