From \(3a^2+3b^2=10ab\)

\(\Rightarrow3a^2+3b^2-10ab=0\)

\(\Rightarrow3a^2-9ab-ab+3b^2=0\)

\(\Rightarrow3a\left(a-3b\right)-b\left(a-3b\right)=0\)

\(\Rightarrow\left(3a-b\right)\left(a-3b\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}3a-b=0\\a-3b=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}b=3a\\a=3b\end{matrix}\right.\)

*)\(b=3a\Rightarrow P=\dfrac{a-b}{a+b}=\dfrac{a-3a}{a+3a}=\dfrac{-2}{4}=-\dfrac{1}{2}\)

*)\(a=3b\Rightarrow P=\dfrac{a-b}{a+b}=\dfrac{3b-b}{3b+b}=\dfrac{2}{4}=\dfrac{1}{2}\)

\(3a^2+3b^2=10ab\Leftrightarrow3a^2-10ab+3b^2=0\)

\(\Rightarrow3a^2-9ab-ab+3b^2=0\Rightarrow3a\left(a-3b\right)-b\left(a-3b\right)=0\)

\(\Rightarrow\left(3a-b\right)\left(a-3b\right)=0\)\(\Rightarrow\left[{}\begin{matrix}3a-b=0\\a-3b=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a=\dfrac{1}{3}b\\a=3b\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}P=\dfrac{\dfrac{1}{3}b-b}{\dfrac{1}{3}b+b}=\dfrac{-\dfrac{2}{3}}{\dfrac{4}{3}}=-\dfrac{1}{2}\\P=\dfrac{3b-b}{3b+b}=\dfrac{2}{4}=\dfrac{1}{2}\end{matrix}\right.\)

So,\(P=\pm\dfrac{1}{2}\)