`21, y=\sqrt{x^2-4}`
ĐKXĐ: `x^2-4>=0`
`<=> (x-2)(x+2)>=0`
`<=>`\(\left[ \begin{array}{l}x≥2\\x≤-2\end{array} \right.\)
Vậy `D=(-oo;-2]∪[2;+oo)`
`22, y=\sqrt{3-4x-4x^2}`
ĐKXĐ: `3-4x-4x^2>=0`
`<=> 3-6x+2x-4x^2>=0`
`<=> (2-x)(3+2x)>=0`
`<=> (2x+3)(x-2)<=0`
Do `2x+3>x-2`
`<=> 2x+3>=0>=x-2`
`<=> -3/2<=x<=2`
Vậy `D=[-3/2;2]`
`23, y=1/\sqrt{2+x}`
ĐKXĐ: `2+x>0<=>x> -2`
Vậy `D=(-2;+oo)`
`24, y=(2x-1)/\sqrt{(x-3)^2}`
ĐKXĐ: `(x-3)^2>0`
`<=> x-3\ne0`
`<=> x\ne3`
Vậy `D=RR\\{3}`
`25, y=\root[3]{x-1}+\sqrt{x^2+2}`
Ta thấy `x^2+2>0` với `AAx`
Vậy `D=RR`
`26, y=x/\root[3]{x^2-4}-2/\sqrt{4x^2-1}`
ĐKXĐ: `{(x^2-4\ne0),(4x^2-1>0):}`
`<=> {(x^2\ne4),((2x-1)(2x+1)>0):}`
`<=>`$\begin{cases}x\ne±2\\\left[ \begin{array}{l}x≥\dfrac{1}{2}\\x≤\dfrac{-1}{2}\end{array} \right.\end{cases}$
Vậy `D=(-oo;-1/2]∪[1/2;+oo)\\{2;-2}`