设实数x.y满足y+x^2=0,若0
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设实数x.y满足y+x^2=0,若0设实数x.y满足y+x^2=0,若0设实数x.y满足y+x^2=0,若0首先y=-x^2代入不等式,即有:要证明loga[a^x+a^(-x^2)]2a^(1/8)
设实数x.y满足y+x^2=0,若0
设实数x.y满足y+x^2=0,若0
设实数x.y满足y+x^2=0,若0
首先y=-x^2代入不等式,
即有:要证明 loga[a^x+a^(-x^2)]2a^(1/8)
而:a^x+a^(-x^2)≥2a^[(x-x^2)/2] (1)(用公式)
对于函数(x-x^2)/2来说,易得其最大值是当x=1/2时,值为1/8.
明显2a^[(x-x^2)/2]≥2a^1/8 ,
所以a^x+a^(-x^2)≥2a^(1/8),
所以loga[a^x+a^(-x^2)]≤loga2 +1/8
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