Giải:

Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x+y-3}{z}=\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{1}{x+y+z}=\frac{x+y-3+y+z+1+x+z+2}{x+y+z}=\frac{2x+2y+2z}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)

\(\Rightarrow\frac{1}{x+y+z}=2\)

\(\Rightarrow x+y+z=\frac{1}{2}\)

Xét \(\frac{x+y-3}{z}=2\)

\(\Rightarrow x+y-3=2z\)

\(\Rightarrow x+y+z-3=3z\)

\(\Rightarrow\frac{1}{2}-3=3z\)

\(\Rightarrow\frac{-5}{2}=3z\)

\(\Rightarrow z=\frac{-5}{6}\)

Xét \(\frac{y+z+1}{x}=2\)

\(\Rightarrow y+z+1=2x\)

\(\Rightarrow x+y+z+1=3x\)

\(\Rightarrow\frac{1}{2}+1=3x\)

\(\Rightarrow\frac{3}{2}=3x\)

\(\Rightarrow x=\frac{1}{2}\)

Xét \(\frac{x+z+2}{y}=2\)

\(\Rightarrow x+z+2=2y\)

\(\Rightarrow x+y+z+2=3y\)

\(\Rightarrow\frac{1}{2}+2=3y\)

\(\Rightarrow\frac{5}{2}=3y\)

\(\Rightarrow y=\frac{5}{6}\)

Vậy bộ số \(\left(x;y;z\right)\) là \(\left(\frac{1}{2};\frac{5}{6};\frac{-5}{6}\right)\)

\(x,y,z>0\)

Áp dụng BĐT Caushy cho 3 số ta có:

\(x^3+y^3+z^3\ge3\sqrt[3]{x^3y^3z^3}=3xyz\ge3.1=3\)

\(P=\dfrac{x^3-1}{x^2+y+z}+\dfrac{y^3-1}{x+y^2+z}+\dfrac{z^3-1}{x+y+z^2}\)

\(=\dfrac{\left(x^3-1\right)^2}{\left(x^2+y+z\right)\left(x^3-1\right)}+\dfrac{\left(y^3-1\right)^2}{\left(x+y^2+z\right)\left(y^3-1\right)}+\dfrac{\left(z^3-1\right)^2}{\left(x+y+z^2\right)\left(x^3-1\right)}\)

Áp dụng BĐT Caushy-Schwarz ta có:

\(P\ge\dfrac{\left(x^3+y^3+z^3-3\right)^2}{\left(x^2+y+z\right)\left(x^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)}\)

\(\ge\dfrac{\left(3-3\right)^2}{\left(x^2+y+z\right)\left(x^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)}=0\)

\(P=0\Leftrightarrow x=y=z=1\)

Vậy \(P_{min}=0\)

TH1:x+y+z=0 \(\Rightarrow x=y=z=0\)

TH2:x+y+z\(\ne0\)

Áp dụng t/c .............

Được x+y+z=1/2

Biến đổi ta được \(x=\frac{1}{2};y=\frac{1}{2};z=-\frac{1}{2}\)