Đáp án: $TXD:D = R\backslash \left\{ { - 1;2;\dfrac{{ - 1 \pm \sqrt {33} }}{2}} \right\}$
Giải thích các bước giải:
$\begin{array}{l}
a)y = \dfrac{{x - \sqrt[3]{{x + 7}}}}{{\sqrt {{x^2} - 6x + 9} - \left| {{x^2} - 5} \right|}}\\
Dkxd:\left\{ \begin{array}{l}
{x^2} - 6x + 9 \ge 0\\
\sqrt {{x^2} - 6x + 9} - \left| {{x^2} - 5} \right| \ne 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{\left( {x - 3} \right)^2} \ge 0\left( {tm} \right)\\
\sqrt {{{\left( {x - 3} \right)}^2}} \ne \left| {{x^2} - 5} \right|
\end{array} \right.\\
\Leftrightarrow \left| {x - 3} \right| \ne \left| {{x^2} - 5} \right|\\
\Leftrightarrow \left\{ \begin{array}{l}
x - 3 \ne {x^2} - 5\\
x - 3 \ne - {x^2} + 5
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{x^2} - x - 2 \ne 0\\
{x^2} + x - 8 \ne 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left( {x - 2} \right)\left( {x + 1} \right) \ne 0\\
x \ne \dfrac{{ - 1 \pm \sqrt {33} }}{2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ne 2\\
x \ne - 1\\
x \ne \dfrac{{ - 1 \pm \sqrt {33} }}{2}
\end{array} \right.\\
Vậy\,TXD:D = R\backslash \left\{ { - 1;2;\dfrac{{ - 1 \pm \sqrt {33} }}{2}} \right\}
\end{array}$
