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【题目】函数fx对一切实数x,y均有fx+y-fyx+2y+1x成立,且f1=0

1求f0的值;

2求fx的解析式

【答案】见解析

【解析】1fx+y-fyx+2y+1x,

令x=1,y=0,得f1-f0=2

f1=0,f0=-2

2令y=0,得fx-f0x+1x

fx=x2+x-2

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