tim x,y biet 43xy5chia het cho 45 y toan 6 23/08/2021 Bởi Anna tim x,y biet 43xy5chia het cho 45 y toan 6
Đáp án: \(\left( {x;y} \right) \in \left\{ \matrix{ \left( {0;6} \right);\left( {6;0} \right);\left( {1;5} \right);\left( {5;1} \right);\left( {2;4} \right);\left( {4;2} \right);\left( {3;3} \right) \hfill \cr \left( {6;9} \right);\left( {9;6} \right);\left( {7;8} \right);\left( {8;7} \right) \hfill \cr} \right\}\) Giải thích các bước giải: \(\overline {43xy5} \) có tận cùng là 5 \( \Rightarrow \overline {43xy5} \,\, \vdots \,\,5\). Để \(\overline {43xy5} \,\, \vdots \,\,45\) thì \(\overline {43xy5} \,\, \vdots \,\,9\) \(\begin{array}{l} \Rightarrow \left( {4 + 3 + x + y + 5} \right)\,\, \vdots \,\,9\\ \Rightarrow \left( {12 + x + y} \right)\,\, \vdots \,\,9\end{array}\) Vì \(0 \le x,\,\,y \le 9 \Rightarrow 0 \le x + y \le 18\) \( \Rightarrow x + y \in \left\{ {6;15} \right\}\). TH1: \(x + y = 6\). \( \Rightarrow \left( {x;y} \right) \in \left\{ {\left( {0;6} \right);\left( {6;0} \right);\left( {1;5} \right);\left( {5;1} \right);\left( {2;4} \right);\left( {4;2} \right);\left( {3;3} \right)} \right\}\) TH2: \(x + y = 15\). \( \Rightarrow \left( {x;y} \right) \in \left\{ {\left( {6;9} \right);\left( {9;6} \right);\left( {7;8} \right);\left( {8;7} \right)} \right\}\). Vậy \(\left( {x;y} \right) \in \left\{ \matrix{ \left( {0;6} \right);\left( {6;0} \right);\left( {1;5} \right);\left( {5;1} \right);\left( {2;4} \right);\left( {4;2} \right);\left( {3;3} \right) \hfill \cr \left( {6;9} \right);\left( {9;6} \right);\left( {7;8} \right);\left( {8;7} \right) \hfill \cr} \right\}\) Bình luận
Đáp án:
\(\left( {x;y} \right) \in \left\{ \matrix{
\left( {0;6} \right);\left( {6;0} \right);\left( {1;5} \right);\left( {5;1} \right);\left( {2;4} \right);\left( {4;2} \right);\left( {3;3} \right) \hfill \cr
\left( {6;9} \right);\left( {9;6} \right);\left( {7;8} \right);\left( {8;7} \right) \hfill \cr} \right\}\)
Giải thích các bước giải:
\(\overline {43xy5} \) có tận cùng là 5 \( \Rightarrow \overline {43xy5} \,\, \vdots \,\,5\).
Để \(\overline {43xy5} \,\, \vdots \,\,45\) thì \(\overline {43xy5} \,\, \vdots \,\,9\)
\(\begin{array}{l} \Rightarrow \left( {4 + 3 + x + y + 5} \right)\,\, \vdots \,\,9\\ \Rightarrow \left( {12 + x + y} \right)\,\, \vdots \,\,9\end{array}\)
Vì \(0 \le x,\,\,y \le 9 \Rightarrow 0 \le x + y \le 18\)
\( \Rightarrow x + y \in \left\{ {6;15} \right\}\).
TH1: \(x + y = 6\).
\( \Rightarrow \left( {x;y} \right) \in \left\{ {\left( {0;6} \right);\left( {6;0} \right);\left( {1;5} \right);\left( {5;1} \right);\left( {2;4} \right);\left( {4;2} \right);\left( {3;3} \right)} \right\}\)
TH2: \(x + y = 15\).
\( \Rightarrow \left( {x;y} \right) \in \left\{ {\left( {6;9} \right);\left( {9;6} \right);\left( {7;8} \right);\left( {8;7} \right)} \right\}\).
Vậy \(\left( {x;y} \right) \in \left\{ \matrix{
\left( {0;6} \right);\left( {6;0} \right);\left( {1;5} \right);\left( {5;1} \right);\left( {2;4} \right);\left( {4;2} \right);\left( {3;3} \right) \hfill \cr
\left( {6;9} \right);\left( {9;6} \right);\left( {7;8} \right);\left( {8;7} \right) \hfill \cr} \right\}\)