2 câu trả lời
Đáp án:
min=-1, max=1/7 Lời giải: \({1^2} + {\left( { - 1} \right)^2} < {\left( { - 3} \right)^2} \Rightarrow \sin x - \cos x + 3 \ne 0\,\,\forall x\) \(\eqalign{ & y = {{\sin x + \cos x - 1} \over {\sin x - \cos x + 3}} \cr & \Leftrightarrow y\sin x - y\cos x + 3y = \sin x + \cos x - 1 \cr & \Leftrightarrow \left( {y - 1} \right)\sin x - \left( {y + 1} \right)\cos x = - 3y - 1 \cr & PT\,co\,\,nghiem \cr & \Leftrightarrow {\left( {y - 1} \right)^2} + {\left( {y + 1} \right)^2} \ge {\left( {3y + 1} \right)^2} \cr & \Leftrightarrow 2{y^2} + 2 \ge 9{y^2} + 6y + 1 \cr & \Leftrightarrow 7{y^2} + 6y - 1 \le 0 \cr & \Leftrightarrow - 1 \le y \le {1 \over 7} \cr & \Rightarrow \min y = - 1 \cr & \max y = {1 \over 7} \cr} \)
Đáp án:
$\max y = \dfrac{1}{7};\,\,\,\min y = - 1$
Lời giải:
\(\begin{array}{l} y = \dfrac{{\sin x + \cos x - 1}}{{\sin x - \cos x + 3}}\\ \Leftrightarrow y\left( {\sin x - \cos x + 3} \right) = \sin x + \cos x - 1\\ \Leftrightarrow \left( {y - 1} \right)\sin x - \left( {y + 1} \right)\cos x = - 3y - 1\,\,\left( 1 \right)\\ \Rightarrow \text{Phương trình }\left( 1 \right)\text{ có nghiệm }\Leftrightarrow {\left( {y - 1} \right)^2} + {\left( {y + 1} \right)^2} \ge {\left( { - 3y - 1} \right)^2}\\ \Leftrightarrow 2{y^2} + 2 \ge 9{y^2} + 6y + 1\\ \Leftrightarrow 7{y^2} + 6y - 1 \le 0\\ \Leftrightarrow - 1 \le y \le \dfrac{1}{7}\\ \Rightarrow \max y = \dfrac{1}{7};\,\,\,\min y = - 1. \end{array}\)
