设函数f(x)在(-∞,+∞)内有定义,f(0)不等于0,f(xy)=f(x)f(y),证明:f(x)=1 已知函数f(x)的定义域为(0,+∞),当x>1时,f(x)...

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\uff081\uff09\u2235f\uff08x?y\uff09=f\uff08x\uff09+f\uff08y\uff09\uff0c\u2234f\uff081\uff09=f\uff081\u00d71\uff09=f\uff081\uff09+f\uff081\uff09=2f\uff081\uff09\uff0c\u2234f\uff081\uff09=0\uff0e\u8bbex 1 \uff0cx 2 \u2208\uff080\uff0c+\u221e\uff09\u4e14x 1 \uff1cx 2 \uff0c\u5219 x 2 x 1 \uff1e1\uff0c\u2234f\uff08 x 2 x 1 \uff09\uff1e0\uff0c\u2234f\uff08x 1 \uff09-f\uff08x 2 \uff09=f\uff08x 1 \uff09-f\uff08 x 2 x 1 ?x 1 \uff09=f\uff08x 1 \uff09-f\uff08 x 2 x 1 \uff09-f\uff08x 1 \uff09=-f\uff08 x 2 x 1 \uff09\uff1c0\u2234f\uff08x 1 \uff09\uff1cf\uff08x 2 \uff09\uff0c\u2234f\uff08x\uff09\u5728\uff080\uff0c+\u221e\uff09\u4e0a\u662f\u589e\u51fd\u6570\uff0e\uff082\uff09\u4ee4x= 1 3 \uff0cy=1\u5f97\uff0cf\uff08 1 3 \u00d71\uff09=f\uff08 1 3 \uff09+f\uff081\uff09\uff0c\u2234f\uff081\uff09=0\uff0e\u4ee4x=3\uff0cy= 1 3 \u5f97\uff0cf\uff081\uff09=f\uff083\u00d7 1 3 \uff09=f\uff083\uff09+f\uff08 1 3 \uff09\uff0c\u2235f\uff08 1 3 \uff09=-1\uff0c\u2234f\uff083\uff09=1\uff0e\u4ee4x=y=3\u5f97\uff0cf\uff089\uff09=f\uff083\uff09+f\uff083\uff09=2\uff0c\u2234f\uff08x\uff09-f\uff08 1 x-2 \uff09\uff1ef\uff089\uff09\uff0cf\uff08x\uff09\uff1ef\uff08 9 x-2 \uff09\u2234 x\uff1e0 x-2\uff1e0 x(x-2)\uff1e9 \uff0c\u89e3\u5f97x\uff1e1+ 10 \uff0e\u2234x\u7684\u53d6\u503c\u8303\u56f4\u4e3a\uff081+ 10 \uff0c+\u221e\uff09

令x=y=0
f(0)=f(0)×f(0)
f(0)不等于0,
f(0)=1
令y=0
f(0)=f(x)×f(0)
f(x)=1

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