???

P.An hở

Vì \(x+y+z=0\Rightarrow x+y=-z\)

Ta có: \(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)\)(suy ra từ hằng đẳng thức)

nên  \(x^3+y^3+z^3=\left(x+y\right)^3-3xy\left(x+y\right)+z^3=\left(-z\right)^3-3xy\left(-z\right)+z^3=3xyz\)

Do đó: \(x^3+y^3+z^3=3xyz\left(\text{*}\right)\)

Thay  \(\left(\text{*}\right)\) vào  \(P\), ta được:

\(P=\frac{3xyz}{xyz}=3\)

Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Leftrightarrow\hept{\begin{cases}\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\\\frac{1}{y}+\frac{1}{z}=-\frac{1}{x}\\\frac{1}{x}+\frac{1}{z}=-\frac{1}{y}\end{cases}}\) (*)

Ta có: \(A=\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}\)

\(=\frac{x}{z}+\frac{y}{z}+\frac{x}{y}+\frac{x}{y}+\frac{y}{x}+\frac{z}{x}\)

\(=\left(\frac{x}{z}+\frac{x}{y}\right)+\left(\frac{y}{x}+\frac{y}{z}\right)+\left(\frac{z}{x}+\frac{z}{y}\right)\)

\(=x\left(\frac{1}{z}+\frac{1}{y}\right)+y\left(\frac{1}{x}+\frac{1}{z}\right)+z\left(\frac{1}{x}+\frac{1}{y}\right)\)

Thay (*) vào,ta có : \(A=x.\left(\frac{-1}{x}\right)+y.\left(-\frac{1}{y}\right)+z.\left(-\frac{1}{z}\right)=\left(-1\right)+\left(-1\right)+\left(-1\right)=-3\)

Ta có: x+y+z=0

Suy ra: x+y=-z; y+z=-x; z+x=-y

ta có: \(\left(\frac{x}{y}+1\right)\left(\frac{y}{z}+1\right)\left(\frac{z}{x}+1\right)\)\(=\frac{x+y}{y}.\frac{y+z}{z}.\frac{z+x}{x}\)

                                                                  \(=\frac{-z}{y}.\frac{-x}{z}.\frac{-y}{x}\)

                                                                   \(=-1\)