Question

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

Answer 1

解：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

Answer 2

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(

Answer 3

解：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

Answer 4

原式=x(x+y)(x-y-x-y)=-2xy(x+y)=1*1=1