Toán Tìm tập xác định y = căn 2 + cos x / 1 – sin 2x 15/09/2021 By Ayla Tìm tập xác định y = căn 2 + cos x / 1 – sin 2x
$y=\dfrac{\sqrt2 + \cos x}{1-\sin 2x}$ ĐK: $1-\sin 2x\neq 0$ $\Leftrightarrow \sin 2x\neq 1$ $\Leftrightarrow 2x\neq \dfrac{\pi}{2}+k2\pi$ $\Leftrightarrow x\neq \dfrac{\pi}{4}+k\pi$ $D=\mathbb{R}$ \ $\{\dfrac{\pi}{4}+k\pi\}$ Trả lời
$\begin{array}{l} y = \sqrt {\frac{{2 + \cos x}}{{1 – \sin 2x}}} \\ DK:\frac{{2 + \cos x}}{{1 – \sin 2x}} \ge 0\,\left( * \right)\\ Ta\,co:\,2 + \cos x > 0,\forall x\,(do\,\cos x \ge – 1\,nen\,2 + \cos x \ge 2 – 1 = 1)\\ Khi\,do\,\left( * \right) \Leftrightarrow 1 – \sin 2x > 0 \Leftrightarrow \sin 2x < 1 \Leftrightarrow \sin 2x \ne 1\,\left( {do\,\sin 2x \le 1,\forall x} \right)\\ \Leftrightarrow 2x \ne \frac{\pi }{2} + k2\pi \Leftrightarrow x \ne \frac{\pi }{4} + k\pi \end{array}$ Trả lời
$y=\dfrac{\sqrt2 + \cos x}{1-\sin 2x}$
ĐK: $1-\sin 2x\neq 0$
$\Leftrightarrow \sin 2x\neq 1$
$\Leftrightarrow 2x\neq \dfrac{\pi}{2}+k2\pi$
$\Leftrightarrow x\neq \dfrac{\pi}{4}+k\pi$
$D=\mathbb{R}$ \ $\{\dfrac{\pi}{4}+k\pi\}$
$\begin{array}{l}
y = \sqrt {\frac{{2 + \cos x}}{{1 – \sin 2x}}} \\
DK:\frac{{2 + \cos x}}{{1 – \sin 2x}} \ge 0\,\left( * \right)\\
Ta\,co:\,2 + \cos x > 0,\forall x\,(do\,\cos x \ge – 1\,nen\,2 + \cos x \ge 2 – 1 = 1)\\
Khi\,do\,\left( * \right) \Leftrightarrow 1 – \sin 2x > 0 \Leftrightarrow \sin 2x < 1 \Leftrightarrow \sin 2x \ne 1\,\left( {do\,\sin 2x \le 1,\forall x} \right)\\ \Leftrightarrow 2x \ne \frac{\pi }{2} + k2\pi \Leftrightarrow x \ne \frac{\pi }{4} + k\pi \end{array}$