Đáp án:
$\begin{array}{l}
Do:\left\{ \begin{array}{l}
{\left( {a + 1} \right)^2} \ge 0\forall a;\\
\left| b \right| \ge 0\forall b
\end{array} \right.\\
\Rightarrow {\left( {a + 1} \right)^2} + \left| b \right| < 2\\
\Rightarrow \left[ \begin{array}{l}
{\left( {a + 1} \right)^2} + \left| b \right| = 1\\
{\left( {a + 1} \right)^2} + \left| b \right| = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
{\left( {a + 1} \right)^2} = 0\\
\left| b \right| = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
{\left( {a + 1} \right)^2} = 1\\
\left| b \right| = 0
\end{array} \right.\\
\left\{ \begin{array}{l}
{\left( {a + 1} \right)^2} = 0\\
\left| b \right| = 0
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
a = - 1\\
b = \pm 1
\end{array} \right.\\
a = 0;b = 0\\
a = - 2;b = 0\\
a = - 1;b = 0
\end{array} \right.\\
\Rightarrow \left( {a;b} \right) = \left( { - 1; \pm 1} \right);\left( {0;0} \right);\left( { - 2;0} \right);\left( { - 1;0} \right)
\end{array}$
