mn giúp mk vs cos$^{2019}$x+ sin$^{2020}$x= 1 11/09/2021 Bởi Raelynn mn giúp mk vs cos$^{2019}$x+ sin$^{2020}$x= 1
Đáp án: Giải thích các bước giải: \[\begin{array}{l} {\cos ^{2019}}x + {\sin ^{2020}}x = 1\\ \Leftrightarrow {\cos ^{2019}}x + {\sin ^{2020}}x = {\cos ^2}x + {\sin ^2}x\\ \Leftrightarrow {\cos ^2}x({\cos ^{2017}}x – 1) = {\sin ^2}x(1 – {\sin ^{2018}}x)(*)\\ – 1 \le \cos x \le 1 \Leftrightarrow – 1 \le {\cos ^{2017}}x \le 1 \Rightarrow {\cos ^{2017}}x – 1 \le 0;{\cos ^2}x \ge 0 \Rightarrow VT \le 0\\ – 1 \le \sin x \le 1 \Leftrightarrow 0 \le {\sin ^{2018}}x \le 1 \Rightarrow 1 – {\sin ^{2018}}x \ge 0;{\sin ^2}x \ge 0 \Rightarrow VP \ge 0\\ \Rightarrow (*) \Leftrightarrow \left\{ \begin{array}{l} VP = 0\\ VT = 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} {\sin ^2}x(1 – {\sin ^{2018}}x) = 0\\ {\cos ^2}x({\cos ^{2017}}x – 1) = 0 \end{array} \right. \Leftrightarrow x = \frac{{k\pi }}{2} \end{array}\] Bình luận
Đáp án:
Giải thích các bước giải:
\[\begin{array}{l}
{\cos ^{2019}}x + {\sin ^{2020}}x = 1\\
\Leftrightarrow {\cos ^{2019}}x + {\sin ^{2020}}x = {\cos ^2}x + {\sin ^2}x\\
\Leftrightarrow {\cos ^2}x({\cos ^{2017}}x – 1) = {\sin ^2}x(1 – {\sin ^{2018}}x)(*)\\
– 1 \le \cos x \le 1 \Leftrightarrow – 1 \le {\cos ^{2017}}x \le 1 \Rightarrow {\cos ^{2017}}x – 1 \le 0;{\cos ^2}x \ge 0 \Rightarrow VT \le 0\\
– 1 \le \sin x \le 1 \Leftrightarrow 0 \le {\sin ^{2018}}x \le 1 \Rightarrow 1 – {\sin ^{2018}}x \ge 0;{\sin ^2}x \ge 0 \Rightarrow VP \ge 0\\
\Rightarrow (*) \Leftrightarrow \left\{ \begin{array}{l}
VP = 0\\
VT = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{\sin ^2}x(1 – {\sin ^{2018}}x) = 0\\
{\cos ^2}x({\cos ^{2017}}x – 1) = 0
\end{array} \right. \Leftrightarrow x = \frac{{k\pi }}{2}
\end{array}\]