设函数Y=FX 是定义在(0,+无穷)上的函数 并且满足F(XY)=F(X)+F(Y) F(1/3)=1 当x>1时 F(X)<0 已知fx 是定义在(0,+∞)上的函数,且满足f（xy）=f...

\u8bbe\u51fd\u6570\u662fy\uff1dfx\u662f\u5b9a\u4e49\u5728(0\uff0c+\u221e)\u7684\u51cf\u51fd\u6570 \u5e76\u4e14\u6ee1\u8db3f(xy)\uff1df(x)+f(y) f(\u4e09\u5206\u4e4b\u4e00

\u4ee4x=y=1
\u5219xy=1
f\uff08xy\uff09=f\uff08x\uff09+f\uff08y\uff09
\u6240\u4ee5f(1)=f(1)+f(1)
f(1)=0

f(x)+f(2-x)<2
f(x)+f(y)=f(xy)
\u6240\u4ee5f(x)+f(2-x)=f[x(2-x)]
f(1/3)=1
2=f(1/3)+f(1/3)=f(1/3*1/3)
\u6240\u4ee5f[x(2-x)]<f(1/3*1/3)
f\uff08x\uff09\u662f\u5b9a\u4e49\u5728\uff080\uff0c+\u221e)\u4e0a\u7684\u51cf\u51fd\u6570
\u6240\u4ee5x(2-x)>1/3*1/3
x^2-2x+1/9<0
9x^2-18x+1<0
\u6240\u4ee5(3-2\u221a2)/3<x<(3+2\u221a2)/3
\u4e14\u6709\u5b9a\u4e49\u57dfx>0
\u6240\u4ee5(3-2\u221a2)/3<x<(3+2\u221a2)/3

\u539f\u9898\u5e94\u8be5\u662f\u4e0d\u662f\u8fd9\u6837\u7684
\u8bbe\u51fd\u6570Y=F(X)\u662f\u5b9a\u4e49\u5728(0,+\u221e)\u4e0a\u7684\u51cf\u51fd\u6570,\u5e76\u4e14\u6ee1\u8db3F(XY)=F(X)+F(Y)\uff0cf(1/3)=1


1\uff09\u6c42f(1)\u7684\u503c
2\uff09\u82e5\u5b58\u5728\u5b9e\u6570m\uff0c\u4f7f\u5f97f\uff08m\uff09=2 \u6c42m\u7684\u503c
3\uff09\u5982\u679cf\uff08x\uff09+f\uff082-x\uff09\uff1c2 \uff0c\u6c42x\u7684\u53d6\u503c\u8303\u56f4
\u89e3\u6cd5\u5982\u4e0b\uff1a\u89e3\u7b54\uff1a
1\u9898 F(XY)=F(X)+F(Y) \u628ax=1\u4ee3\u5165 \u5f97f(1)=0
2\u9898 f\uff08m\uff09=f(1/3)+f(1/3)=f(1/9) \u6240\u4ee5 m=1/9
3\u9898 f\uff08x\uff09+f\uff082-x\uff09=f[x*(2-x)]<2=f(1/9)
\u56e0\u4e3a\u662f\u51cf\u51fd\u6570\u6240\u4ee5 x*(2-x)>1/9 \u89e3\u76840<x<2

F(XY)=F(X)+F(Y)
任取0<x1<x2,则x2/x1>1
∴f(x2)=f(x1*x2/x1)=f(x1)+f(x2/x1)
∴f(x2)-f(x1)=f(x2/x1)
∵当x>1时 F(X)<0
∴F(x2/x1)<0
即f(x2)-f(x1)<0
∴f(x1)>f(x2)
∴F(x)在(0,正无穷)是减函数

第2问解不等式用F(1/3)=1

证明:设x1>x2>0;则有:x1/x2>1;f(x1/x2)<0
又f(x1)=f(x1/x2*x2)=f(x1/x2)加f(x2)
即f(x1)-f(x2)=f(x1/x2)<0
即:f(x1)<f(x2)
故f(x)在(0;正无穷)上是减函数

扩展阅读：f x 怎么求 ... 已知函数f x x3 ... f x 与f x+1 的关系 ... 设函数f x x 2 ... fx x是什么函数 ... 函数图像生成器app ... 设函数fx连续且f'(0)大于0 ... fx f x ... 函数 ...