giải giúp mk vs tìm GTLN, GTNN của hàm số sau: a/ y= 2sinx + cosx - 1 b/ y=$2sin^{2}x$ + 3sinxcosx + $5cos^{2}x$ c/ y= $\frac{2sinx+ cosx}{sinx + 2cosx + 4}$

Đáp án:

$\eqalign{ & a \min y = - \sqrt 5 - 1;\,\,\max y = \sqrt 5 - 1 \cr & b)\min y = - {{3\sqrt 2 } \over 2} + {7 \over 2};\,\,\max y = {{3\sqrt 2 } \over 2} + {7 \over 2} \cr & c)\min y = {{ - 4 - \sqrt {71} } \over {11}};\,\,\max y = {{ - 4 + \sqrt {71} } \over {11}} \cr} $

Lời giải:

$\eqalign{ & a)\,\,y = 2\sin x + \cos x - 1 \cr & y = \sqrt 5 \left( {{2 \over {\sqrt 5 }}\sin x + {1 \over {\sqrt 5 }}\cos x} \right) - 1 \cr & y = \sqrt 5 \sin \left( {x + \alpha } \right) - 1 \cr & \Rightarrow - \sqrt 5 - 1 \le y \le \sqrt 5 - 1 \cr & \text{Vậy } \min y = - \sqrt 5 - 1;\,\,\max y = \sqrt 5 - 1 \cr & b)\,\,y = 2{\sin ^2}x + 3\sin x\cos x + 5{\cos ^2}x \cr & y = 1 - \cos 2x + {3 \over 2}\sin 2x + 5.{{1 + \cos 2x} \over 2} \cr & y = {3 \over 2}\sin 2x + {3 \over 2}\cos 2x + {7 \over 2} \cr & y = {3 \over 2}\left( {\sin 2x + \cos 2x} \right) + {7 \over 2} \cr & y = {3 \over 2}\sqrt 2 \sin \left( {2x + {\pi \over 4}} \right) + {7 \over 2} \cr & \Rightarrow - {{3\sqrt 2 } \over 2} + {7 \over 2} \le y \le {{3\sqrt 2 } \over 2} + {7 \over 2} \cr & \text{Vậy } \min y = - {{3\sqrt 2 } \over 2} + {7 \over 2};\,\,\max y = {{3\sqrt 2 } \over 2} + {7 \over 2} \cr & c)\,\,y = {{2\sin x + \cos x} \over {\sin x + 2\cos x + 4}} \cr & \Leftrightarrow 2\sin x + \cos x = y\sin x + 2y\cos x + 4y \cr & \Leftrightarrow \left( {2 - y} \right)\sin x + \left( {1 - 2y} \right)\cos x = 4y \cr & \text{Phương trình có nghiệm } \Leftrightarrow {\left( {2 - y} \right)^2} + {\left( {1 - 2y} \right)^2} \ge {\left( {4y} \right)^2} \cr & \Leftrightarrow 4 - 4y + {y^2} + 1 - 4y + 4{y^2} - 16{y^2} \ge 0 \cr & \Leftrightarrow - 11{y^2} - 8y + 5 \ge 0 \cr & \Leftrightarrow {{ - 4 - \sqrt {71} } \over {11}} \le y \le {{ - 4 + \sqrt {71} } \over {11}} \cr & \text{Vậy } \min y = {{ - 4 - \sqrt {71} } \over {11}};\,\,\max y = {{ - 4 + \sqrt {71} } \over {11}} \cr} $