|
|
$1)$ đặt $a = {4^{{{\sin }^2}x}} + {4^{co{s^2}x}},a > 0$ ta có :
$\begin{array}{l}
\,\,\,\,\,\,\,\left( {6{y^2} + 2y} \right)a = 25{y^2} + 6y + 1\\
\Leftrightarrow \left( {25 - 6a} \right){y^2} + 2\left( {3 - a} \right)y + 1 = 0
\end{array}$
• Với $a = \frac{{25}}{6}$ ta có $2\left( {3 - \frac{{25}}{6}} \right)y + 1 = 0\,\,\,
\Leftrightarrow \,\,\,y = \frac{3}{7}$
Khi đó $y\,\sin \,x = \frac{3}{7}\sin \,x = {\log _2}\left| {\frac{{3\sin x}}{{16}}} \right| \le {\log
_2}\frac{3}{{16}} < - 2$
$ \Rightarrow \,\,\sin \,x < - \frac{7}{3}$ vô lý.
$a \ne \frac{{25}}{6}:\,\,\,\,\,\Delta ' = {a^2} - 16 \ge 0\,\,\,\, \Leftrightarrow \left[
\begin{array}{l}
a \ge 4\,\,\,\,\,\,\left( {a \ne \frac{{25}}{6}} \right)\\
a \le - 4\,\,\,\,(l)
\end{array} \right.$
Suy ra : ${y_{1,2}} = \frac{{a - 3 \pm \sqrt {{a^2} - 16} }}{{25 - 6a}}$
Hệ phương trình trở thành:
$\left\{ \begin{array}{l}
y\,\sin \,x = {\log _2}\left| {\frac{{y\,\sin \,x}}{{1 + 3y}}} \right|\\
y = \frac{{{{2.4}^{{{\sin }^2}x}} - 3}}{{25 - 6\left( {{4^{{{\sin }^2}x}} + {4^{co{s^2}x}}}
\right)}}
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(I)$
Hoặc $\left\{ \begin{array}{l}
y\,\sin \,x = {\log _2}\left| {\frac{{y\,\sin \,x}}{{1 + 3y}}} \right|\\
y = \frac{{{{2.4}^{co{s^2}x}} - 3}}{{25 - 6\left( {{4^{{{\sin }^2}x}} + {4^{co{s^2}x}}}
\right)}}
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(II)$
Xét ($I$) :
$\begin{array}{l}
{\log _2}\left| {\frac{{y\,\sin \,x}}{{1 + 3y}}} \right| = {\log _2}\left| {\sin \,x} \right| - {\log
_2}\left| {3 + \frac{1}{y}} \right|\\
= {\log _2}\left| {\sin \,x} \right| - {\log _2}\left| {3 + \frac{{25 - 6\left( {{4^{{{\sin }^2}x}} +
{4^{co{s^2}x}}} \right)}}{{{{2.4}^{{{\sin }^2}x}} - 3}}} \right|\\
= {\log _2}\left| {\sin \,x} \right| - {\log _2}{2.4^{co{s^2}x}} = {\log _2}\left| {\sin \,x} \right|
- 1 - {\log _2}{4^{co{s^2}x}} \le - 1
\end{array}$
Suy ra : $y\,\sin \,x \le - 1$
Mặt khác do $\left| y \right| \le 1\,\, \Rightarrow \,\,\,\left| {y\,\sin \,x} \right| \le 1\,\,\,\,
\Rightarrow \,\,\,y\,\sin \,x \ge - 1$
Từ đó suy ra : $\left\{ \begin{array}{l}
y\,\sin \,x = - 1\\
{\log _2}\left| {\sin \,x} \right| - 2co{s^2}x = 0
\end{array} \right.$
Vì $\begin{array}{l}
{\log _2}\left| {\sin \,x} \right| \le 0\\
- 2co{s^2}x \le 0
\end{array}$ nên hệ tương đương với :
$\left\{ \begin{array}{l}
y\,\sin \,x = - 1\\
{\log _2}\left| {\sin \,x} \right| = 0\\
cos\,x = 0
\end{array} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}
\left| {\sin \,x} \right| = 1\\
y\,\sin \,x = - 1
\end{array} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\left[ \begin{array}{l}
\left\{ \begin{array}{l}
\sin \,x = - 1\\
y = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
\sin \,x = 1\\
y = - 1
\end{array} \right.
\end{array} \right.$
Thử lại :
- Với $\left\{ \begin{array}{l}
\sin \,x = - 1\\
y = 1
\end{array} \right.$ thế vào hệ ta có: $\left\{ \begin{array}{l}
- 1 = {\log _2}\frac{1}{4}\\
8.5 = 32
\end{array} \right.$
Hệ không thỏa mãn .
- Với $\left\{ \begin{array}{l}
\sin \,x = 1\\
y = - 1
\end{array} \right.$ ta có: $\left\{ \begin{array}{l}
- 1 = {\log _2}\frac{1}{2}\\
4.5= 20
\end{array} \right.$ hệ thỏa mãn.
Ta có: $\left\{ \begin{array}{l}
\sin \,x = 1\\
y = - 1
\end{array} \right.$
Xét $(II)$ : Giải tương tự ta có: $\left\{ \begin{array}{l}
y\,\sin \,x = - 1\\
{\log _2}\left| {\sin \,x} \right| = 0\\
\sin x = 0
\end{array} \right.$
Hệ này vô nghiệm.
Vậy : nghiệm của hệ là:
$\left\{ \begin{array}{l}
x = k\pi \\
y = 1
\end{array} \right.,\,\,\,\,\,\,\,\,\,\,\,\,k \in Z\,\,\,\,\,\,\,$
|
|
|
Đăng bài 08-05-12 04:17 PM
|
|