Đáp án: `D=\mathbb{R} \ \backslash \ {\frac{π}{2}+kπ, \frac{π}{4}+kπ | k∈\mathbb{Z}}`
`y=\frac{tanx}{1-tanx}`
Đkxđ:
$\begin{cases} cosx \ne 0 ⇔ x \ne \dfrac{\pi}{2}+kπ \\ 1-tanx \ne 0 ⇔ tanx \ne 1 ⇔ x \ne \dfrac{\pi}{4}+kπ \end{cases}$
Vậy `D=\mathbb{R} \ \backslash \ {\frac{π}{2}+kπ, \frac{π}{4}+kπ | k∈\mathbb{Z}}`