已知函数f(x)的定义域为R,对任意实数m,n都有f(m+n)=f(m)*f(n),且当x>0时,0

已知函数f(x)的定义域为R,对任意实数m,n都有f(m+n)=f(m)*f(n),且当x>0时,0
已知函数f(x)的定义域为R,对任意实数m,n都有f(m+n)=f(m)*f(n),且当x>0时,0

已知函数f(x)的定义域为R,对任意实数m,n都有f(m+n)=f(m)*f(n),且当x>0时,0
（1）对于任意实数n有f(n)=f(0+n)=f(0)*f(n) ∴f(0)=1
任意实数m有f(m-m)=f(m)*f(-m)=f(0)=1 ∴f(m)=1/f(-m)
当m>0时0

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1、取m=n=0。有f(0)=f(0)*f(0),所以f(0)=0或1。如果f(0)=0，令n=0时，f(m)=0恒成立，显然是不对的所以f(0)=1。
取令m+n=0,并假定m>0。所以f(0)=f(m)*f(-m)=1,f(-m)=1/f(m),因为x>0时，01。
2、设x与x+a,a<0,f(x)-f(x+a)=f(x)[1-f(...

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1、取m=n=0。有f(0)=f(0)*f(0),所以f(0)=0或1。如果f(0)=0，令n=0时，f(m)=0恒成立，显然是不对的所以f(0)=1。
取令m+n=0,并假定m>0。所以f(0)=f(m)*f(-m)=1,f(-m)=1/f(m),因为x>0时，01。
2、设x与x+a,a<0,f(x)-f(x+a)=f(x)[1-f(a)]<0[由上题知f(x)>0,1-f(a)<0],由于x>x+a,所以是减函数。
3、f(x^2)*f(y^2)=f(x^2+y^2)>f(1),由于是减函数，所以x^2+y^2<1[表示圆内的部分]；
f(ax-y+2)=1=f(0),所以ax-y+2=0[表示直线]
由于A∩B=空集，即直线与圆没有交点或只是相切[因为是圆内]。所以圆心到直线的距离>=半径的长度1。
|a*0-0+2|/(a^2+1)^0.5>=1,所以-3^0.5<=a<=3^0.5。
有些地方只是给了大概思路，自己还需要想想！

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