\(a,y=tan\left(\dfrac{x}{2}-\dfrac{\pi}{6}\right)=\dfrac{sin\left(\dfrac{x}{2}-\dfrac{\pi}{6}\right)}{cos\left(\dfrac{x}{2}-\dfrac{\pi}{6}\right)}\\ HSXD:cos\left(\dfrac{x}{2}-\dfrac{\pi}{6}\right)\ne0\\ \Leftrightarrow\dfrac{x}{2}-\dfrac{\pi}{6}\ne\dfrac{\pi}{2}+\dfrac{k2\pi}{2}\\ \Leftrightarrow\dfrac{x}{2}\ne\dfrac{2}{3}\pi+k\pi\\ \Leftrightarrow x\ne\dfrac{4}{3}\pi+k2\pi\\ \Rightarrow D=R\backslash\left\{\dfrac{4\pi}{3}+k2\pi\right\}\)
\(b,y=cot\left(2x-\dfrac{\pi}{4}\right)=\dfrac{cos\left(2x-\dfrac{\pi}{4}\right)}{sin\left(2x-\dfrac{\pi}{4}\right)}\\ HSXD:sin\left(2x-\dfrac{\pi}{4}\right)\ne0\\ \Leftrightarrow2x-\dfrac{\pi}{4}\ne\dfrac{k2\pi}{2}\\ \Leftrightarrow2x\ne\dfrac{\pi}{4}+k\pi\\ \Leftrightarrow x\ne\dfrac{\pi}{8}+\dfrac{k\pi}{2}\\ \Rightarrow D=R\backslash\left\{\dfrac{\pi}{8}+\dfrac{k\pi}{2}\right\}\)
\(c,y=-\dfrac{2}{sin3x}\\ HSXD:sin3x\ne0\\ \Leftrightarrow3x\ne k\pi\\ \Leftrightarrow x\ne\dfrac{k\pi}{3}\\ \Rightarrow D=R\backslash\left\{\dfrac{k\pi}{3}\right\}\)
d: \(0< =cos^2x< =1\)
=>\(0>=-cos^2x>=-1\)
=>\(3>=-cos^2x+3>=-1+3=2\)
vậy: \(-cos^2x+3\ne0\forall x\)
vậy: TXĐ là D=R
e: ĐKXĐ: \(3\cdot sinx-\Omega\ne0\)
=>\(3\cdot sinx\ne\Omega\)
=>\(sinx\ne\dfrac{\Omega}{3}\)(luôn đúng)
Vậy: TXĐ là D=R
