当x=2007,y=2008时,代数式x&sup4-y&sup4/x²-2xy-y² ÷x²+y²/y-x的值为

问题描述:

当x=2007,y=2008时,代数式x&sup4-y&sup4/x&sup2-2xy-y&sup2 ÷x&sup2+y&sup2/y-x的值为
当x=2007,y=2008时,代数式 (x的四次方-y的四次方)/(x²-2xy-y²) ÷(x²+y²)/(y-x)的值为

(x^4-y^4)/(x²-2xy-y²) ÷(x²+y²)/(y-x)
=(x²-y²)(x²+y²)/(x-y)^2÷(x²+y²)/(y-x)
=(x²-y²)*(x²+y²)*(y-x)/(x-y)^2*(x²+y²)
=(x²-y²))/(y-x)
=(x-y)*(x+y)/(y-x)
=--(x+y)
=--x--y
又x=2007,y=2008,所以(x^4-y^4)/(x²-2xy-y²) ÷(x²+y²)/(y-x)=--x--y= --4015