\(y=\dfrac{1+cot(π/3 +x)}{tan^2(3x-π/4)}= \dfrac{1+\dfrac{cos(\pi/3+x)}{sin(\pi/3+x)}}{\dfrac{sin^2(3x-\pi/4)}{cos^2(3x-\pi/4)}}\)
Hàm số xác định tương đương: \(\left\{\begin{matrix}
sin(\pi/3+x) \ne 0\\
sin(3x-\pi/4) \ne 0\\
cos(3x-\pi/4) \ne 0
\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} \pi/3+x\ne k\pi\\ 3x-\pi/4 \ne k\pi\\ 3x-\pi/4 \ne \pi/2+k2\pi \\ 3x-\pi/4 \ne -\pi/2+k2\pi \end{matrix}\right.\)
\(\Leftrightarrow\begin{cases} x \ne kπ - \dfrac{π}{3} \\ x \ne \dfrac{kπ}{3} + \dfrac{π}{12} \\
x \ne \dfrac{\pi}{4}+\dfrac{k2\pi}3 \\
x \ne -\dfrac{\pi}{12} +\dfrac{k2\pi}3
\end{cases}\)
Vậy tập xác định của hàm số là \(D=R \setminus \{ \dfrac{-\pi}{3}+k\pi; \dfrac{\pi}{12}+\dfrac{k\pi}{3};\dfrac{\pi}{4}+\dfrac{k2\pi}3;-\dfrac{\pi}{12} +\dfrac{k2\pi}3 \}\)