\(x^2-12x+4=0\)
Áp dụng định lý Vi-ét ta có :
\(\left\{{}\begin{matrix}x_1+x_2=12\\x_1.x_2=4\end{matrix}\right.\)
Theo đề bài ta có :
\(P=\dfrac{x_1^2+x_2^2}{\sqrt[]{x_1^2x_2}+\sqrt[]{x_1x_2^2}}\)
\(\Leftrightarrow P=\dfrac{\left(x_1+x_2\right)^2-2x_1x_2}{\sqrt[]{x_1x_2}\left(x_1+x_2\right)}\)
\(\Leftrightarrow P=\dfrac{12^2-2.4}{\sqrt[]{4}.12}\)
\(\Leftrightarrow P=\dfrac{144-8}{2.12}\)
\(\Leftrightarrow P=\dfrac{136}{24}=\dfrac{17}{3}\)