Đáp án:
$\begin{array}{l}
Dkxd:\left\{ \begin{array}{l}
a + 1 \ge 0\\
a\left( {a + 1} \right) \ge 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
a \ge - 1\\
\left[ \begin{array}{l}
a \ge 0\\
a \le - 1
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
a = - 1\\
a \ge 0
\end{array} \right.\\
a + 1 = \sqrt {a\left( {a + 1} \right)} \\
\Leftrightarrow a + 1 - \sqrt {a\left( {a + 1} \right)} = 0\\
\Leftrightarrow {\left( {\sqrt {a + 1} } \right)^2} - \sqrt a .\sqrt {a + 1} = 0\\
\Leftrightarrow \sqrt {a + 1} \left( {\sqrt {a + 1} - \sqrt a } \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt {a + 1} = 0\\
\sqrt {a + 1} = \sqrt a
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
a = - 1\left( {tm} \right)\\
a + 1 = a
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
a = - 1\\
1 = 0\left( {ktm} \right)
\end{array} \right.\\
\text{Vậy}\,a = - 1
\end{array}$
