殡仪馆的停尸间:紧急求助一道数学题!!!

来源:百度文库 编辑:高考问答 时间:2024/04/28 07:16:42
当x=1984,y=1916时,计算〔(x^4-y^4)/(x^2-2xy+y^2)〕*〔(y-x)/(x^2+y^2)〕=__________

需要具体的解答过程。谢谢!!!!!

={[(x^2+y^2)(x+y)(x-y)]/(x-y)^2}*[(y-x)/(x^2+y^2)]
约分,
=[(x+y)/(x-y)]*[-(x-y)]
约分,
=-(x+y)
=-(1984+1916)
=-3900

原式=
(x^2+y^2)*(x^2-y^2) y-x
------------------- * -------- (化简)=
(x-y)^2 x^2+y^2

(x+y)*(x-y)
------------ *-(x-y) (化简) =
(x-y)*(x-y)

-(x+y)

原式=[(x^2-y^2)(x^2+y^2)/(x^2-2xy+y^2)]*[(y-x)/(x^2+y^2)]
=[(x-y)(x+y)(x^2+y^2)/(x-y)^2]*[(y-x)/(x^2+y^2)]
约掉(x-y)和(x^2+y^2)后
=-(X+Y)
=-(1984+1916)
=-3900

((x^4-y^4)/(x^2-2xy+y^2))*((y-x)/(x^2+y^2))
(x^2+y^2)(x^2-y^2)( y-x)
=-----------------------------------
(x-y)^2 *(x^2+y^2)
= -y-x

x=1984,y=1916 -----> -y-x=-3900