搜搜考试网求证：log(2,2^x+2^x²)≥7/8,log(1/2,1/(4^x)+1/(4^x²)≤x+
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求证：log(2,2^x+2^x²)≥7/8,log(1/2,1/(4^x)+1/(4^x²)≤x+y-1.
RT.
RT.
由均值不等式,2^x+2^(x^2)>=2√[2^x*2^(x^2)]=2^[(2+x+x^2)/2]
=2^[(1/2)(x+1/2)^2+7/8]>=2^(7/8),
∴2^x+2^(x^2)>=2^(7/8),
∴log(2^x+2^x²)≥7/8.
仿上,1/4^x+1/4^(x^2)>=2√[(1/4)^(x+x^2)]=2*(1/2)^(x+x^2)=(1/2)(x+x^2-1),
∴1/4^x+1/4^(x^2)>=(1/2)^(x+x^2-1),
∴log[1/(4^x)+1/(4^x²)]≤x+x^2-1.(改题了).
=2^[(1/2)(x+1/2)^2+7/8]>=2^(7/8),
∴2^x+2^(x^2)>=2^(7/8),
∴log(2^x+2^x²)≥7/8.
仿上,1/4^x+1/4^(x^2)>=2√[(1/4)^(x+x^2)]=2*(1/2)^(x+x^2)=(1/2)(x+x^2-1),
∴1/4^x+1/4^(x^2)>=(1/2)^(x+x^2-1),
∴log[1/(4^x)+1/(4^x²)]≤x+x^2-1.(改题了).
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