cho x/y+z + y/z+x +z/x+y =1. Tính S=x^2/y+z +y^2/z+x + z^2/x+y

Question

cho x/y+z + y/z+x +z/x+y =1. Tính S=x^2/y+z +y^2/z+x + z^2/x+y

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Julia 1 tháng 2021-11-08T20:45:42+00:00 2 Answers 3 views 0

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    0
    2021-11-08T20:47:37+00:00

    Ta có:

    $\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}=1$

    $\to (x+y+z)(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y})=1.(x+y+z)$

    $\to (x+y+x)(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y})=x+y+z$(*)

    Mà ta có:

    $(x+y+z)(\dfrac{x}{y+z}+\dfrac{x^2}{y+z}+\dfrac{xy}{y+z}+\dfrac{xz}{y+z}$

    $\to = \dfrac{x^2}{y+z}+\dfrac{x(y+z)}{y+z}$(**)

    $\to = \dfrac{x^2}{y+z}+x$

    Tương tự:

    $(x+y+z)\dfrac{y}{z+x}=\dfrac{y^2}{z+x}+y$(***)

    $(x+y+x)\dfrac{x}{x+y}=\dfrac{z^2}{x+y}+z$(****)

    Từ (*),(**),(***),(****),suy ra:

    $\dfrac{x^2}{y+z}+x+\dfrac{y^2}{z+x}+y+\dfrac{z^2}{x+y}=x+y+z$

    $\to \dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}=0$

    Mà:

    $S=\dfrac{x^2}{y+z}+\dfrac{y^2}{x+z}+\dfrac{z^2}{x+y}$

    $\to S=0$

     

    0
    2021-11-08T20:47:37+00:00

    Ta có: `x/(y+z)+y/(x+z)+z/(x+y)=1`

    `⇒(x+y+z)(x/(y+z)+y/(x+z)+z/(x+y))=x+y+z`

    `⇒(x^2+x(y+z))/(y+z)+(y^2+y(x+z))/(x+z)+(z^2+z(x+y))/(x+y)=x+y+z`

    `⇒x^2/(y+z)+x+y^2/(z+x)+y+z^2/(x+y)+z=x+y+z`

    `⇒x^2/(y+z)+y^2/(z+x)+z^2/(x+y)=0`

    Vậy `S=0`

     

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