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