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) 28/11/2021 Bởi Kinsley 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)
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` Bình luận
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: $P=(\dfrac{x}{y}+3)(\dfrac{y}{z}+3)(\dfrac{z}{x}+3)$ $\to P=\dfrac{x+3y}{y}.\dfrac{y+3z}{z}.\dfrac{z+3x}{x}$ $\to P=\dfrac{(x+3y)(y+3z)(z+3x)}{xyz}=64$ Bình luận
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:
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:
$P=(\dfrac{x}{y}+3)(\dfrac{y}{z}+3)(\dfrac{z}{x}+3)$
$\to P=\dfrac{x+3y}{y}.\dfrac{y+3z}{z}.\dfrac{z+3x}{x}$
$\to P=\dfrac{(x+3y)(y+3z)(z+3x)}{xyz}=64$