9. Theo đề ta có :\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Rightarrow\dfrac{yz+xz+xy}{xyz}=0\Rightarrow yz+xz+xy=0\)
Mặt khác :\(x+y+z=1\Rightarrow\left(x+y+z\right)^2=1\Rightarrow x^2+y^2+z^2+2\left(xy+yz+xz\right)=1\)mà \(yz+xz+xy=0\)
Nên \(x^2+y^2+z^2+2.0=1\Rightarrow x^2+y^2+z^2=1\)
Vậy \(x^2+y^2+z^2=1\)
10.\(\dfrac{1}{x}-\dfrac{1}{y}+\dfrac{1}{z}=0\Rightarrow\dfrac{yz-xz+xy}{xyz}=0\Rightarrow xy-xz+xy=0\)
Ta có :\(x-y+z=-1\Rightarrow\left(x-y+z\right)^2=1\Rightarrow x^2+y^2+z^2-2\left(xy+yz-xz\right)=1\)mà \(xy-xz+xy=0\)
Nên :\(x^2+y^2+z^2-2.0=1\Rightarrow x^2+y^2+z^2=1\)
Vậy \(x^2+y^2+z^2=1\)
11.\(\dfrac{1}{x}+\dfrac{1}{y}-\dfrac{1}{z}=0\Rightarrow\dfrac{yz+xz-xy}{xyz}=0\Rightarrow yz+xz-xy=0\)
\(x+y-z=-2\Rightarrow\left(x+y-z\right)^2=4\Rightarrow x^2+y^2+z^2-2\left(yz+xz-xy\right)=4\)mà \(yz+xz-xy=0\)
Nên :\(x^2+y^2+z^2-2.0=4\Rightarrow x^2+y^2+z^2=4\)
Vậy \(x^2+y^2+z^2=4\)
