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a^2x 2(ab)^x-3b^2x>0
(a^x 3b^x)(a^x-b^x)>0
a>0,b>0,a^x 3b^x>0
=>a^x-b^x>0
a^x>b^x
(a/b)^x>1
所以當(dāng)a>b時x>0當(dāng)a(a^x 3b^x)(a^x-b^x)>0這一步好像漏掉了加號,應(yīng)該是(a^x +3b^x)(a^x-b^x)>0
設(shè)f(x)=log(1/2)[a^2x+2(ab)^x-3b^2x+1](a>0,b>0),解不等式f(x)
設(shè)f(x)=log(1/2)[a^2x+2(ab)^x-3b^2x+1](a>0,b>0),解不等式f(x)
數(shù)學(xué)人氣：331 ℃時間：2020-10-01 11:49:12
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