已知实数x,y满足（x+√x2+2002）（y+√y2+2002）=2002,求x2-3xy-4y2-6x-6y+58的值

令：x+√(x²+2002)=P,
y+√(y²+2002)=Q
所以,(P-x)²=x²+2002...(1)
(Q-y)²=y²+2002...(2)
PQ=2002.(3)
由(1)得：P²-2Px=2002=PQ(4)
由(2)得：Q²-2Qy=2002=PQ(5)
由于PQ=2002≠0,P≠0,Q≠0
所以,
由(4)得：P-2x=Q
由(5)得：Q-2y=P
将上二式左右分别相加得：P+Q-2(x+y)=P+Q
所以,x+y=0
x²-3xy-4y²-6x-6y+58
=(x-4y)(x+y)-6(x+y)+58
=(x+y)(x-4y-6)+58
=0+58=58