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[不等式] 求助:解|f(x)|>g(x)等价于解:f(x)>g(x)或f(x)<-g(x)

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shidilin Posted 2015-4-21 22:03 |Read mode
① 解|f(x)|>g(x)型不等式,有如下公式:

等价于解:f(x)>g(x)或f(x)<-g(x)

②疑惑:为何不需要按照右端g(x)的正负,分类讨论?

谢谢!

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test Posted 2015-4-21 22:39

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 Author| shidilin Posted 2015-4-21 23:14
谢谢!
粘贴过来哦
………………………………………………………………
设A={x||f(x)|>g(x)}, B={x| f(x)<-g(x)}, C={x| f(x)>g(x)}.

(1)若k∈A, 则有|f(k)|>g(k),
若f(k)<0, 则f(k)=-|f(k)|<-g(k), 此时k∈B,
若f(k)≥0, 则f(k)=|f(k)|>g(k), 此时k∈C,
因此必有k∈(B∪C),
所以A包含于(B∪C);
(2)若k∈(B∪C),
当k∈B时, 则有f(k)<-g(k), 即-f(k)>g(x), 于是|f(k)|=|-f(k)|≥-f(k)>g(x), 此时k∈A,
当k∈C时, 则有f(k)>g(k), 于是|f(k)|≥f(k)>g(x), 此时k∈A,
因此必有k∈A,
所以(B∪C)包含于A.

综合(1)(2), A=(B∪C).

从而等价性得证.

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test Posted 2015-4-21 23:32
回复 3# shidilin

贴就贴嘛,怎么还改符号……

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 Author| shidilin Posted 2015-4-21 23:38
来个原版;

设A={x| |f(x)|>g(x)}, B={x| f(x)<-g(x)}, C={x| f(x)>g(x)}.

(1)若k∈A, 则有|f(k)|>g(k),
若f(k)<0, 则f(k)=-|f(k)|<-g(k), 此时k∈B,
若f(k)>=0, 则f(k)=|f(k)|>g(k), 此时k∈C,
因此必有k∈(B∪C),
所以A包含于(B∪C);
(2)若k∈(B∪C),
当k∈B时, 则有f(k)<-g(k), 即-f(k)>g(x), 于是|f(k)|=|-f(k)|>=-f(k)>g(x), 此时k∈A,
当k∈C时, 则有f(k)>g(k), 于是|f(k)|>=f(k)>g(x), 此时k∈A,
因此必有k∈A,
所以(B∪C)包含于A.

综合(1)(2), A=(B∪C).
从而等价性得证.

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