Tìm GTLN và GTNN sin^2x/cos^2x – sin2x +1 02/10/2021 Bởi Madelyn Tìm GTLN và GTNN sin^2x/cos^2x – sin2x +1
$$\eqalign{ & y = {{{{\sin }^2}x} \over {{{\cos }^2}x – \sin 2x + 1}} \cr & \Leftrightarrow y = {{{{1 – \cos 2x} \over 2}} \over {{{1 + \cos 2x} \over 2} – \sin 2x + 1}} \cr & \Leftrightarrow y = {{1 – \cos 2x} \over {1 + \cos 2x – 2\sin 2x + 2}} \cr & \Leftrightarrow y + y\cos 2x – 2y\sin 2x + 2y = 1 – \cos 2x \cr & \Leftrightarrow – 2y\sin 2x + \left( {y + 1} \right)\cos 2x = 1 – 3y \cr & PT\,\,co\,\,nghiem \cr & \Leftrightarrow {\left( { – 2y} \right)^2} + {\left( {y + 1} \right)^2} \ge {\left( {1 – 3y} \right)^2} \cr & \Leftrightarrow 4{y^2} + {y^2} + 2y + 1 \ge 9{y^2} – 6y + 1 \cr & \Leftrightarrow – 4{y^2} + 8y \ge 0 \cr & \Leftrightarrow 0 \le y \le 2 \cr & \Rightarrow \min y = 0;\,{\mathop{\rm maxy}\nolimits} = 2 \cr} $$ Bình luận
$$\eqalign{
& y = {{{{\sin }^2}x} \over {{{\cos }^2}x – \sin 2x + 1}} \cr
& \Leftrightarrow y = {{{{1 – \cos 2x} \over 2}} \over {{{1 + \cos 2x} \over 2} – \sin 2x + 1}} \cr
& \Leftrightarrow y = {{1 – \cos 2x} \over {1 + \cos 2x – 2\sin 2x + 2}} \cr
& \Leftrightarrow y + y\cos 2x – 2y\sin 2x + 2y = 1 – \cos 2x \cr
& \Leftrightarrow – 2y\sin 2x + \left( {y + 1} \right)\cos 2x = 1 – 3y \cr
& PT\,\,co\,\,nghiem \cr
& \Leftrightarrow {\left( { – 2y} \right)^2} + {\left( {y + 1} \right)^2} \ge {\left( {1 – 3y} \right)^2} \cr
& \Leftrightarrow 4{y^2} + {y^2} + 2y + 1 \ge 9{y^2} – 6y + 1 \cr
& \Leftrightarrow – 4{y^2} + 8y \ge 0 \cr
& \Leftrightarrow 0 \le y \le 2 \cr
& \Rightarrow \min y = 0;\,{\mathop{\rm maxy}\nolimits} = 2 \cr} $$