Giải và biện luận hệ phương trình sau tùy theo tham số $m$ :
             $\left\{ \begin{array}{l}
{\log _{\frac{1}{{\sqrt 2 }}}}\left( {\frac{{1 - 5co{s^4}3x}}{2}} \right) \le 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)\\
\frac{{1 + \sin \,x}}{{\sin \,2x}} \le m\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)
\end{array} \right.$


$(1)$ $ \Leftrightarrow 1 - 5co{s^4}3x \ge 1\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,co{s^4}3x \le 0$
      $ \Leftrightarrow \,cos3x = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,x = \frac{\pi }{6} + k\frac{\pi }{3},\,\,\,\,\,\,k \in {\rm Z}$
        Do    $\sin \,2x \ne 0\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,x \ne \,\,h\frac{\pi }{2}\,\,\,\,\,\,\,\,\,\,,\,\,\,\,h \in {\rm Z}$
Suy ra  
Đặt $f(x) = \frac{{1 + \sin \,x}}{{\sin \,2x}}$ ta có các trường hợp sau:
    $l$ chẵn  $\left[ \begin{array}{l}
a)\,\,\,f\left( {\frac{\pi }{6} + l\pi } \right) = \frac{{1 + \frac{1}{2}}}{{\frac{{\sqrt 3 }}{2}}} = \sqrt 3 \\
b)\,\,f\left( { - \frac{\pi }{6} + l\pi } \right) = \frac{{1 - \frac{1}{2}}}{{ - \frac{{\sqrt 3 }}{2}}} =  - \frac{1}{{\sqrt 3 }}
\end{array} \right.$
   $l$ lẻ         $\left[ \begin{array}{l}
a)\,\,\,f\left( {\frac{\pi }{6} + l\pi } \right) = \frac{{1 - \frac{1}{2}}}{{\frac{{\sqrt 3 }}{2}}} = \frac{1}{{\sqrt 3 }}\\
b)\,\,f\left( { - \frac{\pi }{6} + l\pi } \right) = \frac{{1 + \frac{1}{2}}}{{ - \frac{{\sqrt 3 }}{2}}} =  - \sqrt 3
\end{array} \right.$
    Biểu diễn các nghiệm trên trục số, ta có:
  
+Cho $m$ thay đổi ta có:
   $m <  - \sqrt 3 $:  không có giá trị của $f(x) \le m$, hệ vô nghiệm
  
$ - \sqrt 3  \le m <  - \frac{1}{{\sqrt 3 }}$:   hệ có $1$ nghiệm  $x =  - \frac{\pi }{6} + l\pi \,\,\,\,,\,l$ lẻ
   $ - \frac{1}{{\sqrt 3 }} \le m < \frac{1}{{\sqrt 3 }}$ :   hệ có $2$ nghiệm  $\left[ \begin{array}{l}
x =  - \frac{\pi }{6} + l\pi  (l\in Z)\\\,\\
x =  - \frac{\pi }{6} + l\pi  (l\in Z)\,\,\,\,\,
\end{array} \right.$

 

Thẻ

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