f(x)=|(ax-1)/x|=|(ax-1)|/|x|f(-x)=|a+1/x|=|(ax+1)/x|=|(ax+1)|/|x|,假设函数f(x)在其定义域上为偶函数,则f(-x)=f(x),即|(ax-1)|/|x|=|(ax+1)|/|x|,即|(ax-1)|=|(ax+1)|,因为去掉绝对值时,ax+1=ax-1明显不成立,所...