tìm a,b ∈ Z ,SAO CHO a)|a|+a=0 b) |a+5| + |b-4|=0 c)|a|+ |b|=3 d) |a|+ |b| ≤ 29/07/2021 Bởi Arianna tìm a,b ∈ Z ,SAO CHO a)|a|+a=0 b) |a+5| + |b-4|=0 c)|a|+ |b|=3 d) |a|+ |b| ≤
Đáp án: $\begin{array}{l}a)\left| a \right| + a = 0\\ \Rightarrow \left| a \right| = – a\\ \Rightarrow a \le 0\\b)\\\left| {a + 5} \right| + \left| {b – 4} \right| = 0\\Do:\left\{ \begin{array}{l}\left| {a + 5} \right| \ge 0\\\left| {b – 4} \right| \ge 0\end{array} \right. \Rightarrow \left\{ \begin{array}{l}\left| {a + 5} \right| = 0\\\left| {b – 4} \right| = 0\end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l}a + 5 = 0\\b – 4 = 0\end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l}a = – 5\\b = 4\end{array} \right.\\c)\left| a \right| + \left| b \right| = 3 = 0 + 3 = 1 + 2\\ \Rightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}\left| a \right| = 0\\\left| b \right| = 3\end{array} \right.\\\left\{ \begin{array}{l}\left| a \right| = 3\\\left| b \right| = 0\end{array} \right.\\\left\{ \begin{array}{l}\left| a \right| = 1\\\left| b \right| = 2\end{array} \right.\\\left\{ \begin{array}{l}\left| a \right| = 2\\\left| b \right| = 1\end{array} \right.\end{array} \right. \Rightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}a = 0\\b = \pm 3\end{array} \right.\\\left\{ \begin{array}{l}a = \pm 3\\b = 0\end{array} \right.\\\left\{ \begin{array}{l}a = \pm 1\\b = \pm 2\end{array} \right.\\\left\{ \begin{array}{l}a = \pm 2\\b = \pm 1\end{array} \right.\end{array} \right.\\ \Rightarrow \left( {a;b} \right) = \left( {0; \pm 3} \right);\left( { \pm 3;0} \right);\left( { \pm 1; \pm 2} \right);\left( { \pm 2; \pm 1} \right)\end{array}$ Ý d thiếu đề Bình luận
Đáp án:
Giải thích các bước giải:
Đáp án:
$\begin{array}{l}
a)\left| a \right| + a = 0\\
\Rightarrow \left| a \right| = – a\\
\Rightarrow a \le 0\\
b)\\
\left| {a + 5} \right| + \left| {b – 4} \right| = 0\\
Do:\left\{ \begin{array}{l}
\left| {a + 5} \right| \ge 0\\
\left| {b – 4} \right| \ge 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\left| {a + 5} \right| = 0\\
\left| {b – 4} \right| = 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
a + 5 = 0\\
b – 4 = 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
a = – 5\\
b = 4
\end{array} \right.\\
c)\left| a \right| + \left| b \right| = 3 = 0 + 3 = 1 + 2\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
\left| a \right| = 0\\
\left| b \right| = 3
\end{array} \right.\\
\left\{ \begin{array}{l}
\left| a \right| = 3\\
\left| b \right| = 0
\end{array} \right.\\
\left\{ \begin{array}{l}
\left| a \right| = 1\\
\left| b \right| = 2
\end{array} \right.\\
\left\{ \begin{array}{l}
\left| a \right| = 2\\
\left| b \right| = 1
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
a = 0\\
b = \pm 3
\end{array} \right.\\
\left\{ \begin{array}{l}
a = \pm 3\\
b = 0
\end{array} \right.\\
\left\{ \begin{array}{l}
a = \pm 1\\
b = \pm 2
\end{array} \right.\\
\left\{ \begin{array}{l}
a = \pm 2\\
b = \pm 1
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left( {a;b} \right) = \left( {0; \pm 3} \right);\left( { \pm 3;0} \right);\left( { \pm 1; \pm 2} \right);\left( { \pm 2; \pm 1} \right)
\end{array}$
Ý d thiếu đề