Tìm tập xác định của hàm số $y = \dfrac{{\sqrt {x + 2} }}{{x\sqrt {{x^2} - 4x + 4} }}$

${\rm{D}} = \left\{ {2;\dfrac{{ - 3 - \sqrt 5 }}{2};\dfrac{{ - 3 + \sqrt 5 }}{2}} \right\}$

${\rm{D}} = \mathbb{R}\backslash \left\{ { - 2;\dfrac{{ - 3 - 2\sqrt 5 }}{2};\dfrac{{ - 3 + 2\sqrt 5 }}{2}} \right\}$

${\rm{D}} = \mathbb{R}\backslash \left\{ {\dfrac{{ - 3 - \sqrt 5 }}{2};\dfrac{{ - 3 + \sqrt 5 }}{2}} \right\}$

${\rm{D}} = \mathbb{R}\backslash \left\{ {2;\dfrac{{ - 3 - \sqrt 5 }}{2};\dfrac{{ - 3 + \sqrt 5 }}{2}} \right\}$

\(D = \left[ { - 1; + \infty } \right)\backslash \left\{ 0 \right\}\)

${\rm{D}} = \mathbb{R}$  

${\rm{D}} = \left[ { - 1; + \infty } \right)$

${\rm{D}} = \left[ { - 1;1} \right)$

\(m \in \left( { - \infty ;\dfrac{3}{2}} \right] \cup \left\{ 2 \right\}\)

\(m \in \left( { - \infty ; - 1} \right] \cup \left\{ 2 \right\}\)

\(m \in \left( { - \infty ;1} \right] \cup \left\{ 3 \right\}\)

\(m \in \left( { - \infty ;1} \right] \cup \left\{ 2 \right\}\)

\(m = 1\)

\(m =  - 1\)

\(m =  - 2\)

\(m = 2\)

\(N\left( {1;2} \right)\)           

\(N\left( {2; - 2} \right)\)

\(N\left( {1; - 2} \right)\)

\(N\left( {3; - 2} \right)\)

\(y = 2{x^2} + 2x + 2\)

\(y = {x^2} - 4x + 6\)

\(y = {x^2} + 2x + 2\)

\(y = {x^2} + 4x + 6\)