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

\(=\dfrac{\left(x+y\right)^2}{x^2+y^2-xy}=\dfrac{4\left(x^2+y^2-xy\right)-3\left(x^2+y^2-2xy\right)}{x^2+y^2-xy}\)

\(=4-\dfrac{3\left(x-y\right)^2}{x^2+y^2-xy}\le4\)

\(P_{max}=4\) khi \(x=y=\dfrac{1}{2}\)

Ta có : 

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)

\(\Leftrightarrow\)\(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^3=0^3\)

\(\Leftrightarrow\)\(\left(\frac{1}{x}\right)^3+\left(\frac{1}{y}\right)^3+\left(\frac{1}{z}\right)^3+3\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{y}+\frac{1}{z}\right)\left(\frac{1}{z}+\frac{1}{x}\right)=0\)

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

Lại có : 

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)

\(\Rightarrow\)\(\hept{\begin{cases}\frac{1}{x}+\frac{1}{y}=\frac{-1}{z}\\\frac{1}{y}+\frac{1}{z}=\frac{-1}{x}\\\frac{1}{z}+\frac{1}{x}=\frac{-1}{y}\end{cases}}\)

\(\Leftrightarrow\)\(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\left(-3\right).\frac{-1}{z}.\frac{-1}{x}.\frac{-1}{y}\)

\(\Leftrightarrow\)\(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\) ( đpcm ) 

Vậy nếu \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\) thì \(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\)

Chúc bạn học tốt ~ 

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Leftrightarrow\frac{1}{x}+\frac{1}{y}=\frac{-1}{z}\)

\(\Rightarrow\left(\frac{1}{x}+\frac{1}{y}\right)^3=\left(-\frac{1}{z}\right)^3\Leftrightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{3}{x^2y}+\frac{3}{xy^2}=-\frac{1}{z^3}\)

\(\Leftrightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{-3}{x^2y}-\frac{3}{xy^2}=\frac{-3}{xy}.\left(\frac{1}{x}+\frac{1}{y}\right)=\frac{-3}{xy}.-\frac{1}{z}=\frac{3}{xyz}\)

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow\frac{1}{x}+\frac{1}{y}=-\frac{1}{z};\frac{1}{x}+\frac{1}{z}=-\frac{1}{y};\frac{1}{y}+\frac{1}{z}=-\frac{1}{x}\)

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

\(=\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{x^2}-\frac{1}{xy}+\frac{1}{y^2}\right)+\left(\frac{1}{x}+\frac{1}{z}\right)\left(\frac{1}{x^2}-\frac{1}{xz}+\frac{1}{z^2}\right)+\left(\frac{1}{y}+\frac{1}{z}\right)\left(\frac{1}{y^2}-\frac{1}{yz}+\frac{1}{z^2}\right)\)

\(=-\frac{1}{z}\left(\frac{1}{x^2}-\frac{1}{xy}+\frac{1}{y^2}\right)-\frac{1}{y}\left(\frac{1}{x^2}-\frac{1}{xz}+\frac{1}{z^2}\right)-\frac{1}{x}\left(\frac{1}{y^2}-\frac{1}{yz}+\frac{1}{z^2}\right)\)

\(=-\frac{1}{x^2z}+\frac{1}{xyz}-\frac{1}{y^2z}-\frac{1}{x^2y}+\frac{1}{xyz}-\frac{1}{yz^2}-\frac{1}{xy^2}+\frac{1}{xyz}-\frac{1}{xz^2}\)

\(=\left(-\frac{1}{x^2z}-\frac{1}{x^2y}\right)+\left(-\frac{1}{xy^2}-\frac{1}{y^2z}\right)+\left(-\frac{1}{xz^2}-\frac{1}{yz^2}\right)+\frac{3}{xyz}\)

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

\(=-\frac{1}{x^2}\cdot-\frac{1}{x}+-\frac{1}{y^2}\cdot-\frac{1}{y}+-\frac{1}{z^2}\cdot-\frac{1}{z}+\frac{3}{xyz}=\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}+\frac{3}{xyz}\)

\(\Rightarrow2\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)=\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}+\frac{3}{xyz}\Rightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\)(đpcm)

biết làm rồi không giải thì thôi không cần