\(y=2x^2+x+1\)

Hàm số trên luôn xác định với mọi \(x\in \mathbb{R}\)

Vậy tập xác định của hàm số trên là \(D={\mathbb{R}}\).

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\cos\pi x\ne-1\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\pi x\ne\pi+k2\pi\)

\(\Leftrightarrow x\ne2k+1\)

Vậy \(\left\{{}\begin{matrix}x\ge0\\x\ne2k+1\left(k\in Z\right)\end{matrix}\right.\)

ĐKXĐ: \(\left\{{}\begin{matrix}sinx\ne0\Rightarrow x\ne k\pi\\2-\dfrac{\sqrt{3}}{sinx}\ge0\left(1\right)\end{matrix}\right.\) 

Xét (1):

\(\Leftrightarrow\dfrac{2sinx-\sqrt{3}}{sinx}\ge0\Leftrightarrow\left[{}\begin{matrix}sinx\ge\dfrac{\sqrt{3}}{2}\\sinx< 0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{\pi}{3}+k2\pi\le x\le\dfrac{2\pi}{3}+k2\pi\\-\pi+k2\pi< x< k2\pi\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}\dfrac{\pi}{3}+k2\pi\le x\le\dfrac{2\pi}{3}+k2\pi\\-\pi+k2\pi< x< k2\pi\end{matrix}\right.\)

ĐKXĐ: \(\sqrt{x^2+2x+2}-\left(x+1\right)\ge0\)

\(\Rightarrow\sqrt{x^2+2x+2}\ge x+1\)

Ta có \(\sqrt{x^2+2x+2}=\sqrt{\left(x+1\right)^2+1}>\sqrt{\left(x+1\right)^2}=\left|x+1\right|\ge x+1\)

\(\Rightarrow\sqrt{x^2+2x+2}-\left(x+1\right)>0\) \(\forall x\)

\(\Rightarrow D=R\)

ĐKXĐ: (tất cả \(k\in Z\))

a. \(sinx-1\ge0\Leftrightarrow sinx\ge1\)

\(\Leftrightarrow sinx=1\Rightarrow x=\dfrac{\pi}{2}+k2\pi\)

b. \(\left\{{}\begin{matrix}\dfrac{1-sinx}{1+sinx}\ge0\left(luôn-đúng\right)\\1+sinx\ne0\end{matrix}\right.\) \(\Leftrightarrow sinx\ne-1\)

\(\Leftrightarrow x\ne-\dfrac{\pi}{2}+k2\pi\)

c. \(sinx\ne0\Leftrightarrow x\ne k\pi\)

ĐKXĐ: \(sinx+cosx>0\)

\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)>0\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)>0\)

\(\Leftrightarrow k2\pi< x+\dfrac{\pi}{4}< \pi+k2\pi\)

\(\Leftrightarrow-\dfrac{\pi}{4}+k2\pi< x< \dfrac{3\pi}{4}+k2\pi\)

\(y=\sqrt{x^2-2x+1}-\sqrt{x^2+2x+1}\)

\(=\sqrt{\left(x-1\right)^2}-\sqrt{\left(x+1\right)^2}\)

\(=\left|x-1\right|-\left|x+1\right|\)

+)Xét \(x< -1\)\(\Rightarrow\begin{cases}x+1< 0\Rightarrow\left|x+1\right|=-\left(x+1\right)=-x-1\\x-1< 0\Rightarrow\left|x-1\right|=-\left(x-1\right)=-x+1\end{cases}\)

\(\Rightarrow y=\left(-x-1\right)-\left(-x+1\right)=2\)

+)Xét \(-1\le x< 1\)\(\Rightarrow\begin{cases}x\ge-1\Rightarrow x+1\ge0\Rightarrow\left|x+1\right|=x+1\\x< 1\Rightarrow x-1< 0\Rightarrow\left|x-1\right|=-\left(x-1\right)=-x+1\end{cases}\)

\(\Rightarrow y=\left(-x+1\right)-\left(x+1\right)=-2x\)

+)Xét \(x\ge1\)\(\Rightarrow\begin{cases}x-1\ge0\Rightarrow\left|x-1\right|=x-1\\x+1\ge0\Rightarrow\left|x+1\right|=x+1\end{cases}\)

\(\Rightarrow y=\left(x-1\right)-\left(x+1\right)=-2\)

Ta thấy:

\(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{x^2-4}}{x+3}=\lim\limits_{x\rightarrow+\infty}\dfrac{-\sqrt{1-\dfrac{4}{x^2}}}{1+\dfrac{3}{x}}=-1\)

\(\Rightarrow y=-1\) là 1 TCN

\(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{x^2-4}}{x+3}=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{1-\dfrac{4}{x^2}}}{1+\dfrac{3}{x}}=1\)

\(\Rightarrow y=1\) là 1 TCN