Trang chủ » Hỏi đáp » Môn Toán N+15_<n-6=. 2n+15:2n+3=. 6n+9:2n+3= 20/11/2024 N+15_<n-6=. 2n+15:2n+3=. 6n+9:2n+3=
Giải đáp: $\begin{array}{l}a)n \in \left\{ { – 15; – 1;3;5;7;9;13;27} \right\}\\b)n \in \left\{ \begin{array}{l} – \dfrac{{15}}{2}; – \dfrac{9}{2}; – \dfrac{7}{2}; – 3; – \dfrac{5}{2}; – 2;\\ – 1; – \dfrac{1}{2};0;\dfrac{1}{2};\dfrac{3}{2};\dfrac{9}{2}\end{array} \right\}\\c)n \in Z\end{array}$ Lời giải và giải thích chi tiết: $\begin{array}{l}a)n + 15 \vdots n – 6\\ \Leftrightarrow n – 6 + 21 \vdots n – 6\\ \Leftrightarrow 21 \vdots n – 6\\ \Leftrightarrow \left( {n – 6} \right) \in U\left( {21} \right)\\ \Leftrightarrow \left( {n – 6} \right) \in \left\{ { – 21; – 7; – 3; – 1;1;3;7;21} \right\}\\ \Leftrightarrow n \in \left\{ { – 15; – 1;3;5;7;9;13;27} \right\}\\b)\\2n + 15 \vdots 2n + 3\\ \Leftrightarrow 2n + 3 + 12 \vdots 2n + 3\\ \Leftrightarrow 12 \vdots 2n + 3\\ \Leftrightarrow \left( {2n + 3} \right) \in \left\{ \begin{array}{l} – 12; – 6; – 4; – 3; – 2; – 1;\\1;2;3;4;6;12\end{array} \right\}\\ \Leftrightarrow 2n \in \left\{ \begin{array}{l} – 15; – 9; – 7; – 6; – 5; – 4;\\ – 2; – 1;0;1;3;9\end{array} \right\}\\ \Leftrightarrow n \in \left\{ \begin{array}{l} – \dfrac{{15}}{2}; – \dfrac{9}{2}; – \dfrac{7}{2}; – 3; – \dfrac{5}{2}; – 2;\\ – 1; – \dfrac{1}{2};0;\dfrac{1}{2};\dfrac{3}{2};\dfrac{9}{2}\end{array} \right\}\\c)\\6n + 9 \vdots 2n + 3\\ \Leftrightarrow 3.\left( {2n + 3} \right) \vdots 2n + 3\left( {tm} \right)\\Vay\,n \in Z\end{array}$ Trả lời
a)n \in \left\{ { – 15; – 1;3;5;7;9;13;27} \right\}\\
b)n \in \left\{ \begin{array}{l}
– \dfrac{{15}}{2}; – \dfrac{9}{2}; – \dfrac{7}{2}; – 3; – \dfrac{5}{2}; – 2;\\
– 1; – \dfrac{1}{2};0;\dfrac{1}{2};\dfrac{3}{2};\dfrac{9}{2}
\end{array} \right\}\\
c)n \in Z
\end{array}$
a)n + 15 \vdots n – 6\\
\Leftrightarrow n – 6 + 21 \vdots n – 6\\
\Leftrightarrow 21 \vdots n – 6\\
\Leftrightarrow \left( {n – 6} \right) \in U\left( {21} \right)\\
\Leftrightarrow \left( {n – 6} \right) \in \left\{ { – 21; – 7; – 3; – 1;1;3;7;21} \right\}\\
\Leftrightarrow n \in \left\{ { – 15; – 1;3;5;7;9;13;27} \right\}\\
b)\\
2n + 15 \vdots 2n + 3\\
\Leftrightarrow 2n + 3 + 12 \vdots 2n + 3\\
\Leftrightarrow 12 \vdots 2n + 3\\
\Leftrightarrow \left( {2n + 3} \right) \in \left\{ \begin{array}{l}
– 12; – 6; – 4; – 3; – 2; – 1;\\
1;2;3;4;6;12
\end{array} \right\}\\
\Leftrightarrow 2n \in \left\{ \begin{array}{l}
– 15; – 9; – 7; – 6; – 5; – 4;\\
– 2; – 1;0;1;3;9
\end{array} \right\}\\
\Leftrightarrow n \in \left\{ \begin{array}{l}
– \dfrac{{15}}{2}; – \dfrac{9}{2}; – \dfrac{7}{2}; – 3; – \dfrac{5}{2}; – 2;\\
– 1; – \dfrac{1}{2};0;\dfrac{1}{2};\dfrac{3}{2};\dfrac{9}{2}
\end{array} \right\}\\
c)\\
6n + 9 \vdots 2n + 3\\
\Leftrightarrow 3.\left( {2n + 3} \right) \vdots 2n + 3\left( {tm} \right)\\
Vay\,n \in Z
\end{array}$