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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 8: Cho x+y+z=1. Chứng minh: x^2+ y^2 +z^2 => 1/ 3

Toán Lớp 8: Cho x+y+z=1. Chứng minh: x^2+ y^2 +z^2 => 1/ 3

Comments ( 2 )

  1. ~rai~
    \(\text{Áp dụng BĐT Bunhiacopxki,ta có:}\\(x^2+y^2+z^2)(1^2+1^2+1^2)\ge (x.1+y.1+z.1)^2\\\Leftrightarrow 3(x^2+y^2+z^2)\ge (x+y+z)^2\\\Leftrightarrow 3(x^2+y^2+z^2)\ge 1^2\\\Leftrightarrow x^2+y^2+z^2\ge \dfrac{1}{3}\\\text{Dấu “=” xảy ra}\Leftrightarrow \dfrac{x}{1}=\dfrac{y}{1}=\dfrac{z}{1}\\\Leftrightarrow x=y=z=\dfrac{1}{3}.\)

  2. Đáp án:
     
    Giải thích các bước giải:
     

    toan-lop-8-cho-y-z-1-chung-minh-2-y-2-z-2-1-3

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222-9+11+12:2*14+14 = ? ( )

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