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y=log2^[1/(x+1)]+log2^(x^2-1)的反函数(过程、、)
人气:371 ℃ 时间:2020-04-18 12:55:21
解答
y=log2^[1/(x+1)]+log2^(x^2-1)=log2^[1/(x+1)*(x^2-1)=log2^[1/(x+1)*(x-1)(x+1)=log2^(x-1)
反函数x=log2^(y-1)
所以y-1=2^x
y=2^x+1
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