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`

Đáp án:

$P = 64$

Giải thích các bước giải:

$\begin{array}{l}+)\quad x +y + z =0\\ \text{Ta được:}\\ \quad \dfrac{x+3y - z}{z} = \dfrac{y+3z-x}{x} = \dfrac{z+3x-y}{y}\\ \to \dfrac{(x+y-z) + 2y}{z} = \dfrac{(y+z-x)+2z}{x} = \dfrac{(z+x-y)+2x}{y}\\ \to \dfrac{-2z +2y}{z} = \dfrac{-2x+2z}{x} = \dfrac{-2y +2x}{y}\\ \to \dfrac{-z+y}{z} = \dfrac{-x+z}{x} = \dfrac{-y+x}{y}\\ \to \dfrac{y}{z} = \dfrac{z}{x} = \dfrac{x}{y}\\ \to x = y= z \\ \to x = y= z = 0\quad (loại)\\ +)\quad x +y + z \ne 0\\ \text{Áp dụng tính chất dãy tỉ số bằng nhau ta được:}\\ \dfrac{x+3y - z}{z} = \dfrac{y+3z-x}{x} = \dfrac{z+3x-y}{y}\\ = \dfrac{x+3y-z+y+3z-x+z+3x-y}{x+y+z} = \dfrac{3(x+y+z)}{x+y+z} = 3\\ \\ \to \begin{cases}x + 3y -z = 3z\\y+3z - x = 3x\\z +3x - y = 3y\end{cases}\\ \to \begin{cases}x +3y = 4z\\y+3z = 4x\\z+3x = 4y\end{cases}\\ \to \begin{cases}\dfrac{x+3y}{y} = \dfrac{4z}{y}\\\dfrac{y+3z}{z} = \dfrac{4x}{z}\\\dfrac{z+3x}{x} = \dfrac{4y}{x}\end{cases}\\ \to \begin{cases}\dfrac xy + 3 = \dfrac{4z}{y}\\\dfrac yz + 3 = \dfrac{4x}{z}\\\dfrac zx + 3 = \dfrac{4y}{x}\end{cases}\\ \to \left(\dfrac xy + 3\right)\left(\dfrac yz + 3\right)\left(\dfrac zx + 3\right) = \dfrac{4z}{y}\cdot\dfrac{4x}{z}\cdot \dfrac{4y}{z}\\ \to P =64 \end{array}$