tìm các số nguyên a b thỏa mãn : (a+1)^2+/b/<2 29/07/2021 Bởi Kylie tìm các số nguyên a b thỏa mãn : (a+1)^2+/b/<2
Đáp án: $\begin{array}{l}Do:\left\{ \begin{array}{l}{\left( {a + 1} \right)^2} \ge 0\forall a;\\\left| b \right| \ge 0\forall b\end{array} \right.\\ \Rightarrow {\left( {a + 1} \right)^2} + \left| b \right| < 2\\ \Rightarrow \left[ \begin{array}{l}{\left( {a + 1} \right)^2} + \left| b \right| = 1\\{\left( {a + 1} \right)^2} + \left| b \right| = 0\end{array} \right.\\ \Rightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}{\left( {a + 1} \right)^2} = 0\\\left| b \right| = 1\end{array} \right.\\\left\{ \begin{array}{l}{\left( {a + 1} \right)^2} = 1\\\left| b \right| = 0\end{array} \right.\\\left\{ \begin{array}{l}{\left( {a + 1} \right)^2} = 0\\\left| b \right| = 0\end{array} \right.\end{array} \right. \Rightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}a = – 1\\b = \pm 1\end{array} \right.\\a = 0;b = 0\\a = – 2;b = 0\\a = – 1;b = 0\end{array} \right.\\ \Rightarrow \left( {a;b} \right) = \left( { – 1; \pm 1} \right);\left( {0;0} \right);\left( { – 2;0} \right);\left( { – 1;0} \right)\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
Do:\left\{ \begin{array}{l}
{\left( {a + 1} \right)^2} \ge 0\forall a;\\
\left| b \right| \ge 0\forall b
\end{array} \right.\\
\Rightarrow {\left( {a + 1} \right)^2} + \left| b \right| < 2\\
\Rightarrow \left[ \begin{array}{l}
{\left( {a + 1} \right)^2} + \left| b \right| = 1\\
{\left( {a + 1} \right)^2} + \left| b \right| = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
{\left( {a + 1} \right)^2} = 0\\
\left| b \right| = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
{\left( {a + 1} \right)^2} = 1\\
\left| b \right| = 0
\end{array} \right.\\
\left\{ \begin{array}{l}
{\left( {a + 1} \right)^2} = 0\\
\left| b \right| = 0
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
a = – 1\\
b = \pm 1
\end{array} \right.\\
a = 0;b = 0\\
a = – 2;b = 0\\
a = – 1;b = 0
\end{array} \right.\\
\Rightarrow \left( {a;b} \right) = \left( { – 1; \pm 1} \right);\left( {0;0} \right);\left( { – 2;0} \right);\left( { – 1;0} \right)
\end{array}$