\[\begin{array}{l} y = \sqrt {1 + \frac{1}{2}{{\cos }^2}x} + \frac{1}{2}\sqrt {5 + 2{{\sin }^2}x} \\ = \sqrt {1 + \frac{1}{2}{{\cos }^2}x} + \frac{1}{2}\sqrt {5 + 2\left( {1 - {{\cos }^2}x} \right)} \\ = \sqrt {\frac{{2 + {{\cos }^2}x}}{2}} + \frac{1}{2}\sqrt {7 - 2{{\cos }^2}x} \\ Dat\,\,\,\cos x = t\,\,\,\,\,\left( { - 1 \le t \le 1} \right).\\ \Rightarrow y = \sqrt {\frac{{{t^2} + 2}}{2}} + \frac{1}{2}\sqrt {7 - 2{t^2}} \\ \Rightarrow y' = \frac{t}{{2\sqrt {\frac{{{t^2} + 2}}{2}} }} + \frac{1}{2}.\frac{{ - 4t}}{{2\sqrt {7 - 2{t^2}} }} = \frac{t}{{\sqrt {2\left( {{t^2} + 2} \right)} }} - \frac{t}{{\sqrt {7 - 2{t^2}} }}\\ \Rightarrow y' = 0 \Leftrightarrow \frac{t}{{\sqrt {2\left( {{t^2} + 2} \right)} }} - \frac{t}{{\sqrt {7 - 2{t^2}} }} = 0\\ \Leftrightarrow t\left( {\frac{1}{{\sqrt {2\left( {{t^2} + 2} \right)} }} - \frac{1}{{\sqrt {7 - 2{t^2}} }}} \right) = 0\\ \Leftrightarrow t\left( {\sqrt {7 - 2{t^2}} - \sqrt {2\left( {{t^2} + 2} \right)} } \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} t = 0\\ \sqrt {7 - 2{t^2}} - \sqrt {2\left( {{t^2} + 2} \right)} = 0 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} t = 0\\ 7 - 2{t^2} = 2{t^2} + 4 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} t = 0\\ 4{t^2} = 3 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} t = 0\\ {t^2} = \frac{3}{4} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} t = 0\\ t = \frac{{\sqrt 3 }}{2}\\ t = - \frac{{\sqrt 3 }}{2} \end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l} y\left( { - 1} \right) = y\left( 1 \right) = \frac{{\sqrt 6 + \sqrt 5 }}{2}\\ y\left( 0 \right) = 1 + \frac{{\sqrt 7 }}{2}\\ y\left( { \pm \frac{{\sqrt 3 }}{2}} \right) = \frac{{\sqrt {22} }}{2} \end{array} \right.. \end{array}\] Em so sánh và chọn GTLN, GTNN trong các giá trị trên nhé.