Ta có: \(P=\dfrac{13}{17-x}\le1=\dfrac{13}{13}\)
Dấu " = " khi \(17-x=13\Rightarrow x=4\)
Vậy \(MAX_P=1\) khi x = 4
\(P_{ln}\Leftrightarrow\left\{{}\begin{matrix}\left(17-x\right)_{nn}\\17-x>0\end{matrix}\right.\Rightarrow x=16\)
\(P_{ln}=\dfrac{13}{17-16}=13\)
khi x=16