设函数y=f(x)(x∈R且x≠0)对定义域内任意的x1x2恒有f(x1 * x2)=f(x1)+f(x2)1 求证 f(1)=f(-1)=02 求证y=f(x)是偶函数3 若f(x)为(0,+∞)上的增函数,解不等式f(x)+f(x-1/2)≤0
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