分析:當(dāng)x=1時(shí),有f(1)×f(3)=1;即f(3)=-
;當(dāng)x=3時(shí),有f(3)×f(5)=1;即f(5)=-5;f[f(5)]=f(-5).當(dāng)x=-1時(shí),有f(-1)×f(1)=1即f(-1)=-
;當(dāng)x=-3時(shí),有f(-3)×f(-1)=1即f(-3)=-5;當(dāng)x=-5時(shí),有f(-5)×f(-3)=1即f(-5)=-
;由此能求出f[f(5)].
解答:解:f(x)×f(x+2)=1;
即當(dāng)x=1時(shí),有f(1)×f(3)=1;即f(3)=-
;
當(dāng)x=3時(shí),有f(3)×f(5)=1;即f(5)=-5;
則f[f(5)]=f(-5)
當(dāng)x=-1時(shí),有f(-1)×f(1)=1即f(-1)=-
;
當(dāng)x=-3時(shí),有f(-3)×f(-1)=1即f(-3)=-5;
當(dāng)x=-5時(shí),有f(-5)×f(-3)=1即f(-5)=-
;
則有f[f(5)]=f(-5)=-
.
故答案為:-
.
點(diǎn)評(píng):本題考查函數(shù)的周期性的應(yīng)用,解題時(shí)要認(rèn)真審題,仔細(xì)解答,注意函數(shù)周期性的靈活運(yùn)用.