已知【x+√(x^2+2002)】【y+√(y^2+2002)】=2002,则x^3-3xy-4y^2-6x-6y+58=

问题描述:

已知【x+√(x^2+2002)】【y+√(y^2+2002)】=2002,则x^3-3xy-4y^2-6x-6y+58=

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...