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$1)$ Điều kiện $x \ge 0$
Do $\left| {\sin \sqrt x \ge 0} \right|$ nên ${3^{\left| {\sin \,\sqrt x } \right|}} \ge 1$
$\left| {\cos x} \right| \le 1$
Suy ra : $(1)\,\,\,\, \Leftrightarrow \,\,\,\left\{ \begin{array}{l}
\sin \,\sqrt x = 0\\
\cos x = \pm 1
\end{array} \right.\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\left\{ \begin{array}{l}
\sqrt x = k\pi \\
x = h\pi
\end{array} \right.\,\,\,\,\,\,\,(k,h \in {\rm Z})$
$ \Leftrightarrow \left\{ \begin{array}{l}
x = {k^2}{\pi ^2}\\
x = h\pi
\end{array} \right.\,\,\,\,\,\,\,\,\, \Leftrightarrow \left\{ \begin{array}{l}
h = {k^2}\pi \\
x = {k^2}\pi
\end{array} \right.(1')\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,\,\,\,\,\
\\h \in {\rm Z} nên (1') \Leftrightarrow \,\,\,\left\{ \begin{array}{l}
k = 0\\
x = 0
\end{array} \right.$
Vậy nghiệm của $(1)$ là $x = 0$
2) $co{s^2}xy + \frac{1}{{co{s^2}xy}} \ge 2 \Leftrightarrow {\log _2}\left( {co{s^2}xy + \frac{1}{{co{s^2}xy}}} \right) \ge 1$
Mặt khác ${y^2} - 2y + 2 = \left( {{y^2} - 2y + 1} \right) + 1 = {\left( {y - 1} \right)^2} \ge 1$
$ \Rightarrow \frac{1}{{{y^2} - 2y + 2}} \le 1$
Do đó $(2)$ $ \Leftrightarrow \left\{ \begin{array}{l}
co{s^2}xy = 1\\
y - 1 = 0
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \left\{ \begin{array}{l}
co{s^2}x = 1\\
y = 1
\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\left\{ \begin{array}{l}
x = k\pi \\
y = 1
\end{array} \right.\,\,\,\,\,\,\,\,\,\,(k \in {\rm Z})$
Nghiệm của $(2)$ là $(k\pi ;1)\,\,\,,\,\,\,\,k \in {\rm Z}$
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Đăng bài 26-04-12 03:39 PM
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