为了使函数y=x3+x2/x+1能成为偶函数,只需补充定义该函数在某一点的函数值,即当x=----时,y=------

(1) m-2+1/(m-2)
=[(m-2)²+1]/(m-2)
=(m²-4+1)/(m-2)
=(m²-3)/(m-2)
(2) x/(x-y)-x/y
=[xy-x(x-y)]/x(x-y)
=(xy-x²+xy)/x(x-y)
=x(2y-x)/x(x-y)
=(2y-x)/(x-y)
(3) 1/(1-x)+1/(1+x)+2/(1+x^2)+4/(1+x^4)
=(1+x+1-x)/(1-x)(1+x)+2/(1+x²)+4/(1+x^4)
=2/(1-x²)+2/(1+x²)+4/(1+x^4)
=2(1+x²+1-x²)/(1+x²)(1-x²)+4/(1+x^4)
=4/(1-x^4)+4/(1+x^4)
=4(1+x^4+1-x^4)/(1+x^4)(1-x^4)
=8/(1-x^8)

y=x3+x2/x+1定义域是x≠-1
x=1,y=x²=1
偶函数
所以补充x=-1,y=1