Cho hàm số: \(y = \dfrac{{mx}}{{\sqrt {x - m + 2}  - 1}}\) với \(m\) là tham số. Tìm \(m\) để hàm số xác định trên \(\left( {0;1} \right)\)

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

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

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

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

${\rm{D}} = \mathbb{R}\backslash \left\{ {0;2} \right\}$

${\rm{D}} = \left[ { - 2; + \infty } \right)$

${\rm{D}} = \left( { - 2; + \infty } \right)\backslash \left\{ {0;2} \right\}$

${\rm{D}} = \left[ { - 2; + \infty } \right)\backslash \left\{ {0;2} \right\}$

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

${\rm{D}} = \mathbb{R}$  

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

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

\(m = 1\)

\(m =  - 1\)

\(m =  - 2\)

\(m = 2\)

\(N\left( {1;2} \right)\)           

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

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

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

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

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

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

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