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数学
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解关于x的不等式log2(x+1)>1/2log2[a(x-2)+1] (a>2)
2和4是底数
人气:350 ℃ 时间:2019-12-07 20:44:07
解答
定义域x+1>0,a(x-2)+1>0,得x>-1,x>2-1/a,所以x>2-1/a
log2(x+1)>1/2log2[a(x-2)+1]
2log2(x+1)>log2[a(x-2)+1]
log2(x+1)²>log2[a(x-2)+1]
(x+1)²>a(x-2)+1
x²+2x+1>ax-2a+1
x²+(2-a)x-2a>0
(x+2a)(x-a)>0
因为a>2,所以x>a或x<-2a
综上:x>a
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