Ta có \(y'=1-\frac{9}{(x-1)^2}=\frac{x^2-2x-8}{(x-1)^2}\) \(y'=0\Leftrightarrow \bigg \lbrack\begin{matrix} x=-2 \ (L)\\ x=4 \end{matrix}\) Ta có} \(y(2)=11;y(4)=7;y(5)=\frac{29}{4}\) Vậy \(\underset{[2;5]}{min}y=7\) khi x = 4; \(\underset{[2;5]}{max}y=11\) khi x =2