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Dao Trong Luan Coordinator

26/11/2017 at 13:53
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2
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Prove that:

\(\left\{{}\begin{matrix}\left|x+y\right|\le\left|x\right|+\left|y\right|\\\left|x-y\right|\ge\left|x\right|-\left|y\right|\end{matrix}\right.\)




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    Nguyễn Hưng Phát 26/11/2017 at 15:42

    The sentences 1:       If |x+y|>|x|+|y|\(\Rightarrow\left(\left|x+y\right|\right)^2>\left(\left|x\right|+\left|y\right|\right)^2\Leftrightarrow x^2+2xy+y^2>x^2+2.\left|x\right|.\left|y\right|+y^2\)

    \(\Rightarrow2xy>2\left|x\right|.\left|y\right|\Leftrightarrow xy>\left|x\right|.\left|y\right|\) (vô lí)\(\Rightarrow\left|x+y\right|\le\left|x\right|+\left|y\right|\)

    The sentences 2:       If |x-y|<|x|-|y|\(\Rightarrow\left(\left|x-y\right|\right)^2< \left(\left|x\right|-\left|y\right|\right)^2\Leftrightarrow x^2-2xy+y^2< x^2-2.\left|x\right|.\left|y\right|+y^2\)

    \(\Rightarrow-2xy< -2.\left|x\right|.\left|y\right|\Leftrightarrow2xy>2.\left|x\right|.\left|y\right|\)\(\Leftrightarrow xy>\left|x\right|.\left|y\right|\)(vô lí) \(\Rightarrow\left|x-y\right|\ge\left|x\right|-\left|y\right|\)

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    Nguyễn Hiền 29/11/2017 at 08:42

    Câu 1: Nếu | x + y |> | x | + | y | ⇒ ( | x + y | ) 2 > ( | x | + | y | ) 2 ⇔ x 2 + 2 x y + y 2 > x 2 + 2 | x | . | y | + Y 2⇒(|x+y|)2>(|x|+|y|)2⇔x2+2xy+y2>x2+2.|x|.|y|+y2

    ⇒ 2 x y > 2 | x | . | y | ⇔ x y > | x | . | y |⇒2xy>2|x|.|y|⇔xy>|x|.|y| (vô lí)  ⇒ | x + y | ≤ | x | + | y |⇒|x+y|≤|x|+|y|

    Câu 2: Nếu | xy | <| x | - | y |  ⇒ ( | x - y | ) 2 < ( | x | - | y | ) 2 ⇔ x 2 - 2 x y + y 2 < x 2 - 2 | x | . | y | + Y 2⇒(|x-y|)2<(|x|-|y|)2⇔x2-2xy+y2<x2-2.|x|.|y|+y2

    ⇒ - 2 x y < - 2. | x | . | y | ⇔ 2 x y > 2. | x | . | y |⇒-2xy<-2.|x|.|y|⇔2xy>2.|x|.|y|⇔ x y > | x | . | y |⇔xy>|x|.|y|(vô lí)   ⇒ | x - y | ≥ | x | - | y |


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