bài 1:a) Viết Ư(-15) b) Viết B(-6)
c)Tìm các số nguyên x thỏa mãn:x thuộc Ư(30)và-5 bé hơn hoặc bằng x bé hơn20
bài 2: Tìm số nguyên n biết:
a)3n+2 chia hết cho n – 1
b*) n2 – n + 3 chia hết cho n – 1
c*) n2 + 5 chia hết cho n +1
d*) 2n – 1 là ước của 5n – 4
bài 3: Tìm các số nguyên x, y biết:
a)(5x -1)(y-4) = 4 b*) xy + x + 2y = 5
$\begin{array}{l} 1)a)Ư\left( { – 15} \right) = \left\{ { – 15; – 5; – 3; – 1;1;3;5;15} \right\}\\ b)B\left( { – 6} \right) = \left\{ { – 6k/k \in Z} \right\}\\ c) – 5 \le x < 20\\ \Rightarrow x \in \left\{ { – 5; – 3; – 1;1;3;5;6;10;15} \right\}\\ 2)a)3n + 2\\ = 3n – 3 + 5\\ = 3\left( {n – 1} \right) + 5\\ Do:3\left( {n – 1} \right) \vdots \left( {n – 1} \right)\\ \Rightarrow 5 \vdots \left( {n – 1} \right)\\ \Rightarrow \left( {n – 1} \right) \in \left\{ { – 5; – 1;1;5} \right\}\\ \Rightarrow n \in \left\{ { – 4;0;2;6} \right\}\\ b){n^2} – n + 3\\ = n\left( {n – 1} \right) + 3\\ \Rightarrow 3 \vdots \left( {n – 1} \right)\\ \Rightarrow \left( {n – 1} \right) \in \left\{ { – 3; – 1;1;3} \right\}\\ \Rightarrow n \in \left\{ { – 2;0;2;4} \right\}\\ d)5n – 4 \vdots \left( {2n – 1} \right)\\ \Rightarrow 2\left( {5n – 4} \right) \vdots \left( {2n – 1} \right)\\ \Rightarrow 10n – 8 \vdots \left( {2n – 1} \right)\\ \Rightarrow 5.\left( {2n – 1} \right) – 3 \vdots \left( {2n – 1} \right)\\ \Rightarrow 3 \vdots \left( {2n – 1} \right)\\ \Rightarrow 2n – 1 \in \left\{ { – 3; – 1;1;3} \right\}\\ \Rightarrow n \in \left\{ { – 1;0;1;2} \right\}\\ B3)a)\left( {5x – 1} \right).\left( {y – 4} \right) = 4\\ \Rightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} 5x – 1 = 1\\ y – 4 = 4 \end{array} \right.\\ \left\{ \begin{array}{l} 5x – 1 = 4\\ y – 4 = 1 \end{array} \right.\\ \left\{ \begin{array}{l} 5x – 1 = – 1\\ y – 4 = – 4 \end{array} \right.\\ \left\{ \begin{array}{l} 5x – 1 = – 4\\ y – 4 = – 1 \end{array} \right.\\ 5x – 1 = y – 4 = 2\\ 5x – 1 = y – 4 = – 2 \end{array} \right. \Rightarrow \left[ \begin{array}{l} x = \dfrac{2}{5}\left( {ktm} \right)\\ x = 1;y = 5\\ x = 0;y = 0\\ x = – \dfrac{3}{5}\left( {ktm} \right)\\ x = \dfrac{3}{5}\left( {ktm} \right)\\ x = – \dfrac{1}{5}\left( {ktm} \right) \end{array} \right.\\ Vậy\,\left( {x;y} \right) = \left\{ {\left( {1;5} \right);\left( {0;0} \right)} \right\}\\ b)xy + x + 2y = 5\\ \Rightarrow x.\left( {y + 1} \right) + 2y + 2 = 7\\ \Rightarrow x\left( {y + 1} \right) + 2\left( {y + 1} \right) = 7\\ \Rightarrow \left( {y + 1} \right)\left( {x + 2} \right) = 7\\ \Rightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} y + 1 = 1\\ x + 2 = 7 \end{array} \right.\\ \left\{ \begin{array}{l} y + 1 = – 1\\ x + 2 = – 7 \end{array} \right.\\ \left\{ \begin{array}{l} y + 1 = 7\\ x + 2 = 1 \end{array} \right.\\ \left\{ \begin{array}{l} y + 1 = – 7\\ x + 2 = – 1 \end{array} \right. \end{array} \right. \Rightarrow \left[ \begin{array}{l} y = 0;x = 5\\ y = – 2;x = – 9\\ y = 6;x = – 1\\ y = – 8;x = – 3 \end{array} \right.\\ Vậy\,\left( {x;y} \right) = \left\{ {\left( {5;0} \right);\left( { – 9; – 2} \right);\left( { – 1;6} \right);\left( { – 3; – 8} \right)} \right\} \end{array}$
Đáp án:
$\begin{array}{l}
1)a)Ư\left( { – 15} \right) = \left\{ { – 15; – 5; – 3; – 1;1;3;5;15} \right\}\\
b)B\left( { – 6} \right) = \left\{ { – 6k/k \in Z} \right\}\\
c) – 5 \le x < 20\\
\Rightarrow x \in \left\{ { – 5; – 3; – 1;1;3;5;6;10;15} \right\}\\
2)a)3n + 2\\
= 3n – 3 + 5\\
= 3\left( {n – 1} \right) + 5\\
Do:3\left( {n – 1} \right) \vdots \left( {n – 1} \right)\\
\Rightarrow 5 \vdots \left( {n – 1} \right)\\
\Rightarrow \left( {n – 1} \right) \in \left\{ { – 5; – 1;1;5} \right\}\\
\Rightarrow n \in \left\{ { – 4;0;2;6} \right\}\\
b){n^2} – n + 3\\
= n\left( {n – 1} \right) + 3\\
\Rightarrow 3 \vdots \left( {n – 1} \right)\\
\Rightarrow \left( {n – 1} \right) \in \left\{ { – 3; – 1;1;3} \right\}\\
\Rightarrow n \in \left\{ { – 2;0;2;4} \right\}\\
d)5n – 4 \vdots \left( {2n – 1} \right)\\
\Rightarrow 2\left( {5n – 4} \right) \vdots \left( {2n – 1} \right)\\
\Rightarrow 10n – 8 \vdots \left( {2n – 1} \right)\\
\Rightarrow 5.\left( {2n – 1} \right) – 3 \vdots \left( {2n – 1} \right)\\
\Rightarrow 3 \vdots \left( {2n – 1} \right)\\
\Rightarrow 2n – 1 \in \left\{ { – 3; – 1;1;3} \right\}\\
\Rightarrow n \in \left\{ { – 1;0;1;2} \right\}\\
B3)a)\left( {5x – 1} \right).\left( {y – 4} \right) = 4\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
5x – 1 = 1\\
y – 4 = 4
\end{array} \right.\\
\left\{ \begin{array}{l}
5x – 1 = 4\\
y – 4 = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
5x – 1 = – 1\\
y – 4 = – 4
\end{array} \right.\\
\left\{ \begin{array}{l}
5x – 1 = – 4\\
y – 4 = – 1
\end{array} \right.\\
5x – 1 = y – 4 = 2\\
5x – 1 = y – 4 = – 2
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = \dfrac{2}{5}\left( {ktm} \right)\\
x = 1;y = 5\\
x = 0;y = 0\\
x = – \dfrac{3}{5}\left( {ktm} \right)\\
x = \dfrac{3}{5}\left( {ktm} \right)\\
x = – \dfrac{1}{5}\left( {ktm} \right)
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left\{ {\left( {1;5} \right);\left( {0;0} \right)} \right\}\\
b)xy + x + 2y = 5\\
\Rightarrow x.\left( {y + 1} \right) + 2y + 2 = 7\\
\Rightarrow x\left( {y + 1} \right) + 2\left( {y + 1} \right) = 7\\
\Rightarrow \left( {y + 1} \right)\left( {x + 2} \right) = 7\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
y + 1 = 1\\
x + 2 = 7
\end{array} \right.\\
\left\{ \begin{array}{l}
y + 1 = – 1\\
x + 2 = – 7
\end{array} \right.\\
\left\{ \begin{array}{l}
y + 1 = 7\\
x + 2 = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
y + 1 = – 7\\
x + 2 = – 1
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
y = 0;x = 5\\
y = – 2;x = – 9\\
y = 6;x = – 1\\
y = – 8;x = – 3
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left\{ {\left( {5;0} \right);\left( { – 9; – 2} \right);\left( { – 1;6} \right);\left( { – 3; – 8} \right)} \right\}
\end{array}$