Đáp án:
\(TXD:D = \left[ { - 9; + \infty } \right)\backslash \left\{ 2 \right\}\)
Giải thích các bước giải:
\(\begin{array}{l}
y = \dfrac{{\sqrt {x + 9} }}{{{x^2} + 8x - 20}}\\
DK:\left\{ \begin{array}{l}
x + 9 \ge 0\\
{x^2} + 8x - 20 \ne 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \ge - 9\\
\left( {x - 2} \right)\left( {x + 10} \right) \ne 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \ge - 9\\
x \ne \left\{ { - 10;2} \right\}
\end{array} \right.\\
\to TXD:D = \left[ { - 9; + \infty } \right)\backslash \left\{ 2 \right\}
\end{array}\)
