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Ta có: $1=x^{2}+y^{2}+z^{2}=(x+y+z)^{2}-2(xy+yz+zx)$nên $P=x+y+z-\frac{(x+y+z)^{2}-1}{2}=\frac{1}{2}(x+y+z+1)^{2}-1 \geq -1 $Khi$x=-1, y=z=0 $thì$P=-1$. Vậy min$p=-1$Ta có:$xy+yz+zx \leq x^{2}+y^{2}+z^{2}=1 $Nên$(x+y+z)^{2}=1+2(xy+yz+zx)\leq 3 $. Do đó$P \leq \sqrt{3}+1 $Khi$x=y=z=\frac{1}{\sqrt{3} } $thì$P=\sqrt{3} $. Vậy max$P=\sqrt{3}+1 $

Ta có: $1=x^{2}+y^{2}+z^{2}=(x+y+z)^{2}-2(xy+yz+zx)$nên $P=x+y+z+\frac{(x+y+z)^{2}-1}{2}=\frac{1}{2}(x+y+z+1)^{2}-1 \geq -1 $Khi$x=-1, y=z=0 $thì$P=-1$. Vậy $\min P=-1$Ta có:$xy+yz+zx \leq x^{2}+y^{2}+z^{2}=1 $Nên$(x+y+z)^{2}=1+2(xy+yz+zx)\leq 3 $. Do đó$P \leq \sqrt{3}+1 $Khi$x=y=z=\frac{1}{\sqrt{3} } $thì$P=\sqrt{3} $. Vậy $\max P=\sqrt{3}+1 $

Ta có: $1=x^{2}+y^{2}+z^{2}=(x+y+z)^{2}-2(xy+yz+zx)$nên $P=x+y+z-\frac{(x+y+z)^{2}-1}{2}=\frac{1}{2}(x+y+z+1)^{2}-1 \geq -1$Khi $x=-1, y=z=0$ thì $P=-1$. Vậy min$p=-1$Ta có: $xy+yz+zx \leq x^{2}+y^{2}+z^{2}=1$Nên $(x+y+z)^{2}=1+2(xy+yz+zx)\leq 3$. Do đó $P \leq \sqrt{3}+1$Khi $x=y=z=\frac{1}{\sqrt{3} }$ thì $P=\sqrt{3}$. Vậy man$P=\sqrt{3}+1$

Ta có: $1=x^{2}+y^{2}+z^{2}=(x+y+z)^{2}-2(xy+yz+zx)$nên $P=x+y+z-\frac{(x+y+z)^{2}-1}{2}=\frac{1}{2}(x+y+z+1)^{2}-1 \geq -1$Khi $x=-1, y=z=0$ thì $P=-1$. Vậy min$p=-1$Ta có: $xy+yz+zx \leq x^{2}+y^{2}+z^{2}=1$Nên $(x+y+z)^{2}=1+2(xy+yz+zx)\leq 3$. Do đó $P \leq \sqrt{3}+1$Khi $x=y=z=\frac{1}{\sqrt{3} }$ thì $P=\sqrt{3}$. Vậy max$P=\sqrt{3}+1$