สรุปให้ครับ
$(x^{2}-y^{2})(x^{2}-y^{2}-2x)=2$
$(x-y)(x+y)(x^{2}-y^{2}-2x)=2$
$(x-y)(x+y)[(x-y)(x+y)-2x]=2$
$\because x-y=1$
$\therefore x=1+y$
$(x+y)(x+y-2x)=2$
$(x+y)(y-x)=2$
$(1+2y)(-1)=2$
$1+2y=-2$
$y=-\frac{3}{2} $
$x=-\frac{1}{2}$
$\therefore xy=\frac{3}{4}$
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