Dễ thấy phương trình có hai nghiệm phân biệt.
Theo định lí Vi-ét: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=10\\x_1x_2=9\end{matrix}\right.\).
Ta có: \(P^2=\dfrac{x_1}{10x_2-9}+\dfrac{x_2}{10x_1-9}+2\sqrt{\dfrac{x_1x_2}{\left(10x_1-9\right)\left(10x_2-9\right)}}\).
\(=\dfrac{x_1\left(10x_1-9\right)+x_2\left(10x_2-9\right)}{\left(10x_1-9\right)\left(10x_2-9\right)}+2\sqrt{\dfrac{x_1x_2}{\left(10x_1-9\right)\left(10x_2-9\right)}}\)
\(=\dfrac{10\left(x_1^2+x_2^2\right)-9\left(x_1+x_2\right)}{100x_1x_2-90\left(x_1+x_2\right)+81}+2\sqrt{\dfrac{x_1x_2}{100x_1x_2-90\left(x_1+x_2\right)+81}}\)
\(=\dfrac{10\left[\left(x_1+x_2\right)^2-2x_1x_2\right]-9\left(x_1+x_2\right)}{100x_1x_2-90\left(x_1+x_2\right)+81}+2\sqrt{\dfrac{x_1x_2}{100x_1x_2-90\left(x_1+x_2\right)+81}}\)
Suy ra: \(P^2=\dfrac{10\left(10^2-2\cdot9\right)-9\cdot10}{100\cdot9-90\cdot10+81}+2\sqrt{\dfrac{9}{100\cdot9-90\cdot10+81}}\)
\(\Rightarrow P^2=\dfrac{784}{81}\Rightarrow P=\dfrac{28}{9}\)
Vậy: \(P=\dfrac{28}{9}.\)
