Cho x, y, z khác 0 và x+3y-z/z=y+3z-x/x=z+3x-y/y tính P= (x/y+3)(y/z+3)(z/x+3)

Question

Cho x, y, z khác 0 và x+3y-z/z=y+3z-x/x=z+3x-y/y
tính P= (x/y+3)(y/z+3)(z/x+3)

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Kinsley 2 tuần 2021-11-28T13:07:29+00:00 2 Answers 12 views 0

Answers ( )

    1
    2021-11-28T13:08:31+00:00

    Đáp án:

    Ta có: `{x+3y-z}/z={y+3z-x}/x={z+3x-y}/y={x+3y-z+y+3z-x+z+3x-y}/{x+y+z}`

            `={(x+y+z)+(3x+3y+3z)-(x+y+z)}/{x+y+z}`

            `={3(x+y+z)}/{x+y+z}=3`

    `=>`$+)$ `{x+3y-z}/z=3`

          `<=>{x+3y}/z-1=3<=>x+3y=4z`

           $+)$ `{y+3z-x}/x=3<=>y+3z=4x`

           $+)$ `{z+3x-y}/y=3<=>z+3x=4y`

    `P=(x/y+3)(y/z+3)(z/x+3)`

      `=({x+3y}/y)({y+3z)/z)({z+3x}/x)`  

      `={4zcdot4xcdot4y}/{xcdotycdotz}={64xyz}/{xyz}=64`

                       Vậy `P=64`

     

    0
    2021-11-28T13:09:06+00:00

    Đáp án: $P=64$

    Giải thích các bước giải:

    Ta có:

    $\dfrac{x+3y-z}{z}=\dfrac{y+3z-x}{x}=\dfrac{z+3x-y}{y}$

    Nếu $x+y+z=0$

    Ta có:

    $ \dfrac{x+3y-z}{z}=\dfrac{y+3z-x}{x}=\dfrac{z+3x-y}{y}$

    $\to \dfrac{x+3y-z}{z}+2=\dfrac{y+3z-x}{x}+2=\dfrac{z+3x-y}{y}+2$

    $\to  \dfrac{x+3y-z+2z}{z}=\dfrac{y+3z-x+2x}{x}=\dfrac{z+3x-y+2y}{y}$

    $\to  \dfrac{x+3y+z}{z}=\dfrac{y+3z+x}{x}=\dfrac{z+3x+y}{y}$

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

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

    $\to  \dfrac{y}{z}=\dfrac{z}{x}=\dfrac{x}{y}$

    $\to  (\dfrac{y}{z})^3=(\dfrac{z}{x})^3=(\dfrac{x}{y})^3=\dfrac{y}z.\dfrac{z}x.\dfrac{x}{y}=1$

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

    $\to x=y=z$

    Mà $x+y+z=0\to x=y=z=0$ (loại )

    Trường hợp $x+y+z\ne 0$

    Ta có:

    $ \dfrac{x+3y-z}{z}=\dfrac{y+3z-x}{x}=\dfrac{z+3x-y}{y}=\dfrac{x+3y-z+y+3z-x+z+3x-y}{z+x+y}$

    $\to \dfrac{x+3y-z}{z}=\dfrac{y+3z-x}{x}=\dfrac{z+3x-y}{y}=\dfrac{3(x+y+z)}{x+y+z}=3$

    $\to \dfrac{x+3y}{z}-1=\dfrac{y+3z}{x}-1=\dfrac{z+3x}{y}-1=3$

    $\to \dfrac{x+3y}{z}=\dfrac{y+3z}{x}=\dfrac{z+3x}{y}=4$

    $\to \dfrac{x+3y}{z}\cdot\dfrac{y+3z}{x}\cdot\dfrac{z+3x}{y}=4^3$

    $\to \dfrac{(x+3y)(y+3z)(z+3x)}{xyz}=64$

    Lại có:

    $P=(\dfrac{x}{y}+3)(\dfrac{y}{z}+3)(\dfrac{z}{x}+3)$

    $\to P=\dfrac{x+3y}{y}.\dfrac{y+3z}{z}.\dfrac{z+3x}{x}$

    $\to P=\dfrac{(x+3y)(y+3z)(z+3x)}{xyz}=64$

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