cho x,y,z thỏa mãn x + y + z=2;x^2 + y^2 + z^2=18;xyz=-1 tính S= 1/(xy + z -1) +1/(xz + y -1) +1/(zy + x - 1)

Ta có : `x+y+z=2`

`<=> x^2+y^2+z^2+2xy+2yz+2zx=4`

`<=> 18+2xy+2yz+2zx=4`

`<=> xy+yz+zx=-7`

Xét `xy+z-1=xy+2-x-y-1=xy-x-y+1=x(y-1)-(y-1)=(y-1)(x-1)`

Tương tự ta có : `xz+y-1=(x-1)(z-1)`

                          `zy+x-1=(z-1)(y-1)`

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

`= (z-1+x-1+y-1)/((x-1)(y-1)(z-1))`

`= (x+y+z-3)/(xyz-(xy+yz+zx)+(x+y+z)-1)`

`= (2-3)/(-1+7+2-1)`

`= -1/7`

Vậy `S=-1/7`