Câu I (3 điểm). 1) Giải phương trình ($\log_{2}\frac{3}{2})^2$ + $2\log_{3}2x$ = $\log_{3}18x$ . $\log_{2}\frac{3}{x}$  2) Giải phương trình  $\frac{cosx}{sin3x}$ + $\frac{cos3x}{sin9x}$ + $\frac{cos9x}{sinx12}$ = 0

Đáp án:

1.$x\in\{\sqrt[5]{\dfrac{27}4}, \dfrac34\}$

2.$x=\dfrac{k\pi}{13}$ hoặc $x=\dfrac{\pi+k2\pi}{26}$

Giải thích các bước giải:

1.ĐKXĐ: $x>0$

Ta có:

$(\log_2\dfrac3x)^2+2\log_32x=\log_318x\cdot\log_2\dfrac3x$

$\to (\log_2\dfrac3x)^2+2\log_32x=\log_33^2\cdot 2x\cdot\log_2\dfrac3x$

$\to (\log_2\dfrac3x)^2+2\log_32x=(2+\log_32x)\cdot\log_2\dfrac3x$

Đặt $\log_2\dfrac3x=a, \log_32x=b$

$\to a^2+2b=(2+b)a$

$\to a^2+2b=2a+ab$

$\to a^2-ab+2b-2a=0$

$\to a(a-b)-2(a-b)=0$

$\to (a-2)(a-b)=0$

$\to a=2$ hoặc $a=b$

Trường hợp $a=2\to \log_2\dfrac3x=2$

$\to \dfrac3x=4$

$\to x=\dfrac34$

Trường hợp $a=b$

$\to \log_2\dfrac3x=\log_32x$

$\to (\log_2\dfrac3x)^6=(\log_32x)^6$

$\to ((\log_2\dfrac3x)^2)^3=((\log_32x)^3)^2$

$\to (\dfrac3x)^3=(2x)^2$

$\to \dfrac{27}{x^3}=4x^2$

$\to x^5=\dfrac{27}4$

$\to x=\sqrt[5]{\dfrac{27}4}$

2.Ta có:

$\dfrac{\cos x}{\sin3x}$

$=\dfrac{\cos x\sin x}{\sin3x\sin x}$

$=\dfrac12\cdot \dfrac{2\cos x\sin x}{\sin3x\sin x}$

$=\dfrac12\dfrac{\sin2x}{\sin3x\sin x}$

$=\dfrac12\dfrac{\sin(3x-x)}{\sin3x\sin x}$

$=\dfrac12\dfrac{\sin3x\cos x-\cos3x\sin x}{\sin3x\sin x}$

$=\dfrac12(\dfrac{\cos x}{\sin x}-\dfrac{\cos3x}{\sin3x})$

$=\dfrac12(\cot x-\cot3x)$

Tương tự:

$\dfrac{\cos3x}{\sin9x}=\dfrac12(\cot3x-\cot9x)$

$\dfrac{\cos9x}{\sin27x}=\dfrac12(\cot9x-\cot27x)$

$\to \dfrac{\cos x}{\sin3x}+\dfrac{\cos3x}{\sin9x}+\dfrac{\cos9x}{\sin27x}=\dfrac12(\cot x-\cot3x)+\dfrac12(\cot3x-\cot9x)+\dfrac12(\cot9x-\cot27x)$

$\to 0=\dfrac12(\cot x-\cot27x)$

$\to \cot x-\cot27x=0$

$\to \cot x=\cot27x$

$\to \dfrac{\cos x}{\sin x}=\dfrac{\cos27x}{\sin27x}$

$\to \cos x\sin27x=\sin x\cos27x$

$\to \sin27x\cos x-\cos27x\sin x=0$

$\to \sin(27x-x)=0$

$\to \sin26x=0$

$\to 26x=0+k2\pi$ hoặc $26x=\pi+k2\pi, k\in Z$

$\to x=\dfrac{k\pi}{13}$ hoặc $x=\dfrac{\pi+k2\pi}{26}$