Ta có: `1/x+1/y+1/z=2`

`=>(1/x+1/y+1/z)^2=4`

`=>1/x^2+1/y^2+1/z^2+2*(1/(xy)+1/(yz)+1/(xz))=4`

`=>1/x^2+1/y^2+1/z^2+2*(1/(xy)+1/(yz)+1/(xz))=2/(xy)-1/z^2`

`=>1/x^2+1/y^2+2/(yz)+2/(xz)+2/z^2=0`

`=>(1/x^2+2/(xz)+1/z^2)+(1/y^2+2/(yz)+1/z^2)=0`

`=>(1/x+1/z)^2+(1/y+1/z)^2=0(**)`

Vì `{((1/x+1/z)^2>=0),((1/y+1/z)^2>=0):}` với mọi `x,y,zne0`

`->(**)` xảy ra `<=>{(1/x+1/z=0),(1/y+1/z=0):}`

`->x=y=-z`

Thay lần lượt `1/x+1/z=0,1/y+1/z=0` vào `1/x+1/y+1/z=2` ta có:

`{(y=1/2),(x=1/2):}`

`->z=-1/2` 

Thay `x,y,z` vào `P` được:

`P=(1/2+2*1/2-1/2)^2024`

`P=1^2024=1`