log1/2[a^2x+(ab)^x-2·b^2x+1]0可以十字相乘：(a^x+2b^x)(a^x-b^x)>0因为a^x+2b^x恒为正,所以：a^x-b^x>0a^x>b^x(a/b)^x>1lg[(a/b)^x]>lg1xlg(a/b)>0当a>b时,a/b>1,则lg(a/b)>0,所以：x>0当a
解不等式log1/2[a^2x+(ab)^x+2·b^2x+1]
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