1. Tìm TXĐ a) y= 3/ cosx – 1 b) tan2x +tan x + 1

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1. a,

   ĐK: $\cos x-1\ne 0$

   $\Leftrightarrow \cos x\ne 1$

   $\Leftrightarrow x\ne k2\pi$

   $\to D=\mathbb{R}$ \ $\{k2\pi\}$

   b, 

   ĐI: $\cos2x\ne 0$, $\cos x\ne 0$

   $\Leftrightarrow x\ne \dfrac{\pi}{4}+k\dfrac{\pi}{2}; x\ne \dfrac{\pi}{2}+k\pi$

   $\to D=\mathbb{R}$ \ $\{\dfrac{\pi}{4}+k\dfrac{\pi}{2}; \dfrac{\pi}{2}+k\pi\}$

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2. Đáp án:

    1.

   a . TXĐ: \(D=R\)\{\(k.2\pi\)} \((k \epsilon Z)\)

   b. TXĐ: \(D=R\)\{\(\dfrac{\pi}{4}+k.\dfrac{\pi}{2};\dfrac{\pi}{2}+k.\pi\)} \((k \epsilon Z)\)

   Giải thích các bước giải:

    1. 

   a. ĐK: \(\cos x-1 \neq 0\)

   \(\Leftrightarrow \cos x \neq 1\)

   \(\Leftrightarrow x \neq k.2\pi\)

   TXĐ: \(D=R\)\{\(k.2\pi\)} \((k \epsilon Z)\)

   b.

   \(\tan 2x+\tan x+1=\dfrac{\sin 2x}{\cos 2x}+\dfrac{\sin x}{\cos x}+1\)

   ĐK: 

   $\begin{cases}\cos 2x \neq 0\\ \cos x \neq 0\end{cases}$

   \(\Leftrightarrow \) $\begin{cases}2x \neq \dfrac{\pi}{2}+k.\pi\\ x \neq \dfrac{\pi}{2}+k.\pi\end{cases}$

   \(\Leftrightarrow \) $\begin{cases}x \neq \dfrac{\pi}{4}+k.\dfrac{\pi}{2}\\ x \neq \dfrac{\pi}{2}+k.\pi\end{cases}$

   TXĐ: \(D=R\)\{\(\dfrac{\pi}{4}+k.\dfrac{\pi}{2};\dfrac{\pi}{2}+k.\pi\)}

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