Tìm TXĐ nha: 1. y=(sin x)/(cos (x-pi)) 2. y=1/(tan x-1) ??? Giúp mình 09/08/2021 Bởi Maya Tìm TXĐ nha: 1. y=(sin x)/(cos (x-pi)) 2. y=1/(tan x-1) ??? Giúp mình
Đáp án: 1. \(D=R\)\{\(\dfrac{3\pi}{2}+k.\pi, k \epsilon Z\)} 2. \(D=R\)\{\(\dfrac{\pi}{4}+k.\pi,\dfrac{\pi}{2}+k.\pi, k \epsilon Z\)} Giải thích các bước giải: 1. \(y=\dfrac{\sin x}{\cos (x-\pi)}\)ĐK: \(\cos (x-\pi) \neq 0\)\(\Leftrightarrow x-\pi \neq \dfrac{\pi}{2}+k.\pi\) \(\Leftrightarrow x \neq \dfrac{3\pi}{2} +k.\pi\) \((k \epsilon Z)\)\(D=R\)\{\(\dfrac{3\pi}{2}+k.\pi, k \epsilon Z\)}2. \(y=\dfrac{1}{\tan x-1}\)ĐK: $\begin{cases}\tan x -1 \neq 0\\\cos x \neq 0 (Do \tan x=\dfrac{\sin x}{\cos x})\end{cases}$\(\Leftrightarrow \) $\begin{cases}\tan x \neq 1\\ x \neq \dfrac{\pi}{2}+k.\pi\end{cases}$ \(\Leftrightarrow \) $\begin{cases}\tan x \neq \tan \dfrac{\pi}{4}\\ x \neq \dfrac{\pi}{2}+k.\pi\end{cases}$ \(\Leftrightarrow \) $\begin{cases}x \neq \dfrac{\pi}{4}+k.\pi\\ x \neq \dfrac{\pi}{2}+k.\pi\end{cases}$ Vậy TXĐ: \(D=R\)\{\(\dfrac{\pi}{4}+k.\pi,\dfrac{\pi}{2}+k.\pi, k \epsilon Z\)} Bình luận
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Đáp án:
1. \(D=R\)\{\(\dfrac{3\pi}{2}+k.\pi, k \epsilon Z\)}
2. \(D=R\)\{\(\dfrac{\pi}{4}+k.\pi,\dfrac{\pi}{2}+k.\pi, k \epsilon Z\)}
Giải thích các bước giải:
1. \(y=\dfrac{\sin x}{\cos (x-\pi)}\)
ĐK: \(\cos (x-\pi) \neq 0\)
\(\Leftrightarrow x-\pi \neq \dfrac{\pi}{2}+k.\pi\)
\(\Leftrightarrow x \neq \dfrac{3\pi}{2} +k.\pi\) \((k \epsilon Z)\)
\(D=R\)\{\(\dfrac{3\pi}{2}+k.\pi, k \epsilon Z\)}
2. \(y=\dfrac{1}{\tan x-1}\)
ĐK: $\begin{cases}\tan x -1 \neq 0\\\cos x \neq 0 (Do \tan x=\dfrac{\sin x}{\cos x})\end{cases}$
\(\Leftrightarrow \) $\begin{cases}\tan x \neq 1\\ x \neq \dfrac{\pi}{2}+k.\pi\end{cases}$
\(\Leftrightarrow \) $\begin{cases}\tan x \neq \tan \dfrac{\pi}{4}\\ x \neq \dfrac{\pi}{2}+k.\pi\end{cases}$
\(\Leftrightarrow \) $\begin{cases}x \neq \dfrac{\pi}{4}+k.\pi\\ x \neq \dfrac{\pi}{2}+k.\pi\end{cases}$
Vậy TXĐ: \(D=R\)\{\(\dfrac{\pi}{4}+k.\pi,\dfrac{\pi}{2}+k.\pi, k \epsilon Z\)}