Question

Tìm TXĐ của hàm số:

\(d,y=\dfrac{2tan\left(x+\dfrac{\pi}{4}\right)}{sinxcosx+cos2x-3}\)

\(e,y=tanx-tan3x\)

Answer 1

d, Hàm số xác định khi:

\(\left\{{}\begin{matrix}cos\left(x+\dfrac{\pi}{4}\right)\ne0\\sinx.cosx+cos2x-3\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{\pi}{4}\ne\dfrac{\pi}{2}+k\pi\\\dfrac{1}{2}sin2x+cos2x\ne3\end{matrix}\right.\)

\(\Leftrightarrow x\ne\dfrac{\pi}{4}+k\pi\)

Answer 2

e, Hàm số xác định khi:

\(\left\{{}\begin{matrix}cosx\ne0\\cos2x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{2}+k\pi\\x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\end{matrix}\right.\)