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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 11: tìm txđ hàm số y= căn sinx+2/2sinxcosx y=1/(2cos^2x-1)tanx

Tháng Một 2, 2023 Toán học 1 Comment 0 views

Toán Lớp 11: tìm txđ hàm số
y= căn sinx+2/2sinxcosx
y=1/(2cos^2x-1)tanx

Comments ( 1 )

  1. bichhhathu
    bichhhathu
    2 Tháng Một, 2023 at 6:13 sáng
    Reply
    ~rai~
    \(1.y=\dfrac{\sqrt{\sin x+2}}{2\sin x\cos x}\\ĐKXĐ:\begin{cases}\sin x+2\ge 0\\2\sin x\cos x\ne 0\end{cases}\quad(1)\\\text{Ta có:}-1\le \sin x\le 1\\\Leftrightarrow 1\le \sin x+2\le 3.\\\Leftrightarrow \sin x+2>0 \quad \forall x\quad nên\\(1)\Leftrightarrow 2\sin x\cos x\ne 0\\\Leftrightarrow \sin 2x\ne 0\\\Leftrightarrow 2x\ne k\pi\\\Leftrightarrow x\ne k\dfrac{\pi}{2}.(k\in\mathbb{Z})\\TXĐ:D=\mathbb{R}\backslash\left\{k\dfrac{\pi}{2}\Big|k\in\mathbb{Z}\right\}.\\2.y=\dfrac{1}{(2\cos^2x-1).\tan x}\\ĐKXĐ:\begin{cases}(2\cos^2x-1).tan x\ne 0\\\cos x\ne 0\end{cases}\\\Leftrightarrow \begin{cases}2\cos^2x-1\ne 0\\\tan x\ne 0\\\cos x\ne 0\end{cases}\\\Leftrightarrow \begin{cases}\cos 2x\ne 0\\\sin x\ne 0\\\cos x\ne 0\end{cases}\\\Leftrightarrow \begin{cases}\cos 2x\ne 0\\2\sin x\cos x\ne 0\end{cases}\\\Leftrightarrow \begin{cases}\cos2x\ne 0\\\sin 2x\ne 0\end{cases}\\\Leftrightarrow 2\sin2x\cos2x\ne 0\\\Leftrightarrow \sin4x\ne 0\\\Leftrightarrow 4x\ne k\pi\\\Leftrightarrow x\ne k\dfrac{\pi}{4}.(k\in\mathbb{Z})\\TXĐ:D=\mathbb{R}\backslash\left\{k\dfrac{\pi}{4}\Big|k\in\mathbb{Z}\right\}.\)

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222-9+11+12:2*14+14 = ? ( )

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