VD1: `y=1/(x^2+4x+5)`
ĐKXĐ: `x^2+4x+5\ne0`
`<=> (x^2+4x+4)+1\ne0`
`<=> (x+2)^2+1\ne0` với `AAx`
Vậy `D=RR`
VD2: `y=(2x+1)/(x^3-3x+2)`
ĐKXĐ: `x^3-3x+2\ne0`
`<=> x^3+2x^2-2x^2-4x+x+2\ne0`
`<=> x^2(x+2)-2x(x+2)+(x+2)\ne0`
`<=> (x+2)(x^2-2x+1)\ne0`
`<=> (x+2)(x-1)^2\ne0`
`<=> {(x+2\ne0),(x-1\ne0):}`
`<=> {(x\ne-2),(x\ne1):}`
Vậy `D=RR\\{-2;1}`
VD3: `y=\sqrt{-2x+3}-\sqrt{x-1}`
ĐKXĐ: `{(-2x+3>=0),(x-1>=0):}`
`<=>`$\begin{cases}2x\le3\\x\ge1\end{cases}$
`<=>`$\begin{cases}x\le\dfrac{3}{2}\\x\ge1\end{cases}$
`<=> 1<=x<=3/2`
Vậy `D=[1;3/2]`
VD4: `y=2/((x+2)\sqrt{x+1})`
ĐKXĐ: `{(x+2\ne0),(x+1>0):}`
`<=> {(x\ne-2),(x> -1):}`
`<=> x> -1`
Vậy `D=(-1;+oo)`
