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HỦY DIỆT THE WORLD

04/12/2017 at 21:05
Answers
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Cho x , y , z , là các số thực thỏa mãn : \(x^2+y^2+z^2=1\)

Tìm mã của p = x + y + z + xy + yz + zx .




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  • ...
    KEITA FC 8C 04/12/2017 at 21:14

    Áp dụng bất đẳng thức Bunhia ta có :

    \(\left(1x+1y+1z\right)^2\)

    \(\le\left(1^2+1^2+1^1\right)\left(x^2+y^2+z^2\right)=3\)

    \(\Rightarrow x+y+z\le\sqrt{3}\)

    \(\left(xy+yz+zx\right)^2\)

    \(\le\left(x^2+y^2+z^2\right)\left(y^2+z^2+x^2\right)=1\)

    \(\Rightarrow xy+yz+zx\le1\)

    \(\Rightarrow p\le1+\sqrt{3}\)

    Mặt khác \(x=y=z=\dfrac{1}{\sqrt{3}}\) thì \(p=1+\sqrt{3}\)

    Vậy maxP \(=1+\sqrt{3}\)

    HỦY DIỆT THE WORLD selected this answer.
  • ...
    Mons Trần 04/12/2017 at 21:29

    You don't speak Vietnamese here.You must speak English

    You are violating rules.

    You should redress.Are you okayy..?


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