已知函数y=f(x)的定义域是R,且满足f(x+y)=f(x)+f(y)-1,f(6)=5,则f(3)
2个回答
解由f(6)=f(3+3)=f(3)+f(3)-1=5
即2f(3)=6
即f(3)=3
故f(9)=f(6+3)=f(6)+f(3)-1=5+3-1=7
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