ILMF

Hoặc:

$\Leftrightarrow \frac{\cos \frac{C}{2}}{\sin \frac{C}{2}}=\frac{\sin C}{\cos (A-B)-\cos C } \Leftrightarrow \cos (A-B)-\cos C=2\sin ^{2}\frac{C}{2} \Leftrightarrow \cos (A-B)-\cos C= 1- \cos C \Leftrightarrow \cos (A-B)=1 \Leftrightarrow A=B$

đpcm

1.

$A=\sin 6^{\circ}.\cos 48^{\circ}.\cos 24^{\circ}.\cos 12^{\circ}= \frac{1}{2\cos 6^{\circ}}.2.\sin 6^{\circ}.\cos 6\dot{^{\circ}}.\cos 48^{\circ}.\cos 24^{\circ}.\cos 12^{\circ}=\frac{1}{2\cos 6^{\circ}}.\sin 12^{\circ}.\cos 12^{\circ}.\cos 48^{\circ}.\cos 24^{\circ}=\frac{1}{4\cos 6^{\circ}}.\sin 24^{\circ}.\cos 24^{\circ}.\cos 48^{\circ}=\frac{1}{8\cos 6^{\circ}}.\sin 48^{\circ}.\cos 48^{\circ}=\frac{1}{16\cos 6^{\circ}}.\sin 96^{\circ}=\frac{1}{16}$

Đây là bài giải:

$\frac{16x}{\sqrt{x+1}} + 8\sqrt{3x} = 9x+21  \Leftrightarrow \frac{16x}{\sqrt{x+1}} -24 +8\sqrt{3x} -24 -9x+27=0\Leftrightarrow (x-3)(\frac{64(4x+3)}{(x+1)(\frac{16x}{\sqrt{x+1}} +24 )}+\frac{1}{8\sqrt{3x}+24}-9)=0\Leftrightarrow x-3=0  \Leftrightarrow x=3$