\(3a^2+2b^2=7ab\)
\(\Leftrightarrow3a^2+2b^2-7ab=0\)
\(\Leftrightarrow3a^2-6ab-ab+2b^2=0\)
\(\Leftrightarrow3a\left(a-2b\right)-b\left(a-2b\right)=0\)
\(\Leftrightarrow\left(3a-b\right)\left(a-2b\right)=0\)
Mà \(3a>b>0\)nên \(3a-b>0\)
Vậy \(a-2b=0\Leftrightarrow a=2b\Leftrightarrow\frac{a}{2}=\frac{b}{1}\)
Đặt \(\frac{a}{2}=\frac{b}{1}=k\Rightarrow\hept{\begin{cases}a=2k\\b=k\end{cases}}\)
\(\Rightarrow P=\frac{2005.2k-2006.k}{2006.2k+2007.k}=\frac{2004k}{6019k}=\frac{2004}{6019}\)