(1)f(x)=log3(x/27)*log3(3x)=(log3(x)-log3(27))*(log3(x)+log3(3))=(log3(x)-3)(log3(x)+1) =(log3(x))^2-2log3(x)-3=(log3(x)-1)^2-4x∈[1/27,1/9],logx∈[-3,-2]故当x=1/27,logx=-3时,f(x)有最大值12当x=1/9,logx=-2时,f(x)有最小值5(2) 令t=log3(x),则方程f(x)+m=0为t^2-2t+m-3=0记方程两根为t1,t2,由韦达定理,t1+t2=2又t1=log3(a),t2=log3(b),故t1+t2=log3(a)+log3(b)=l...
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