问题
填空题
已知x-y=2,x2+y2=4,则x2001+y2001的值为______.
答案
∵x-y=2…①,
x2+y2=4…②,
①2-②得:(x-y)2-(x2+y2)=x2+y2-2xy-x2-y2=22-4=0,
即2xy=0,
∴x=0或y=0,
∵x-y=2,
∴x=0,y=-2或x=2,y=0.
∴x2001+y2001=02001+(-2)2001=-22001
或x2001+y2001=22001+02001=22001
故答案为±22001.