$1.$ Giải phương trình: $2\cos 2x - 8\cos x + 7 = \frac{1}{cos x}$
$2.$ Tam giác $ABC$ có: $\frac{b}{\cos B} + \frac{c}{\cos C} = \frac{a}{\sin B\sin C}$
Chứng minh tam giác $ABC$ là tam giác vuông.
$1.$ Đặt $t = cosx \left( { - 1 \le t \le 1} \right)$
$\begin{array}{l}
PT \Leftrightarrow 2(2{t^2} - 1) - 8t + 7 = \frac{1}{t}\,\,\,(t \ne 0)\\

 \Leftrightarrow \left[ \begin{array}{l}
t = \frac{1}{2}\\
t = 1
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x =  \pm \frac{\pi }{3} + 2k\pi \\
x = 2k\pi
\end{array} \right.
\end{array}$$,k\in Z$
2. $\frac{b}{{\cos B}} + \frac{c}{{\cos C}} = \frac{a}{{\sin B\sin C}}$

$\Leftrightarrow\frac{2R\sin B}{\cos B}+\frac{2R\sin C}{\cos C}=\frac{2R\sin A}{\sin B\sin C}$

$\begin{array}{l}
 \Leftrightarrow \tan B + \tan C = \frac{{\sin A}}{{\sin B\sin C}} \Leftrightarrow \frac{{\sin (B +
C)}}{{\cos B\cos C}} = \frac{{\sin A}}{{\sin B\sin C}}\\
 \Leftrightarrow \cos B\cos C = \sin B\sin C \Leftrightarrow c{\rm{os}}(B + C) = 0\\
 \Leftrightarrow \cos A = 0 \Leftrightarrow A = \frac{\pi }{2}
\end{array}$
Tam giác $ABC$ vuông tại $A.$
cảm ơn lời giải nhé, hay đó –  kellyhoang297 07-11-12 09:40 PM
cám ơn bạn nhé ^_^ –  roilevitinh_hn 06-07-12 10:39 AM

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