1) 17 chia hết cho 2x+3
2) x+12 chia hết cho x+5
3) x-1 chia hết cho x-13
4) 2x+1 là ước của 2x+23
5) x-2 là ước của 2x
6) x+10 chia hết cho 2x-3
7) 2x+1 chia hết cho x+3
8)x-1 là ước của 3x+25
9) 2x+3 chia hết cho 3x+1
10) 5x+4 là ước của của 4x-2
giúp mik nha mik vote cho 5 sao
Đáp án:
9) \(\left[ \begin{array}{l}
x = 2\\
x = 0
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1)17 \vdots 2x + 3\\
\Leftrightarrow 2x + 3 \in U\left( {17} \right)\\
\to \left[ \begin{array}{l}
2x + 3 = 17\\
2x + 3 = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 7\\
x = – 1
\end{array} \right.\\
2)x + 12 \vdots x + 5\\
\to x + 5 + 7 \vdots x + 5\\
\to 7 \vdots x + 5\\
\to x + 5 \in U\left( 7 \right)\\
\to \left[ \begin{array}{l}
x + 5 = 7\\
x + 5 = 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 2\\
x = – 4
\end{array} \right.\\
3)x – 1 \vdots x – 13\\
\to x – 13 + 12 \vdots x – 13\\
\to 12 \vdots x – 13\\
\to x – 13 \in U\left( {12} \right)\\
\to \left[ \begin{array}{l}
x – 13 = 12\\
x – 13 = 6\\
x – 13 = 4\\
x – 13 = 3\\
x – 13 = 2\\
x – 13 = 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 25\\
x = 19\\
x = 17\\
x = 16\\
x = 15\\
x = 14
\end{array} \right.\\
4)2x + 1 \in U\left( {2x + 23} \right)\\
\to 2x + 23 \vdots 2x + 1\\
\to 2x + 1 + 22 \vdots 2x + 1\\
\to 22 \vdots 2x + 1\\
\to 2x + 1 \in U\left( {22} \right)\\
\to \left[ \begin{array}{l}
2x + 1 = 22\\
2x + 1 = 11\\
2x + 1 = 2\\
2x + 1 = 1
\end{array} \right. \to \left[ \begin{array}{l}
x = \dfrac{{21}}{2}\\
x = 5\\
x = \dfrac{1}{2}\\
x = 0
\end{array} \right.\\
5)x – 2 \in U\left( {2x} \right)\\
\to 2x \vdots x – 2\\
\to 2\left( {x – 2} \right) + 4 \vdots x – 2\\
\to 4 \vdots x – 2\\
\to x + 2 \in U\left( 4 \right)\\
\to \left[ \begin{array}{l}
x + 2 = 4\\
x + 2 = 2\\
x + 2 = 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 2\\
x = 0\\
x = – 1
\end{array} \right.\\
6)x + 10 \vdots 2x – 3\\
\to 2x + 20 \vdots 2x – 3\\
\to 2x – 3 + 23 \vdots 2x – 3\\
\to 23 \vdots 2x – 3\\
\to 2x – 3 \in U\left( {23} \right)\\
\to \left[ \begin{array}{l}
2x – 3 = 23\\
2x – 3 = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 13\\
x = 2
\end{array} \right.\\
7)2x + 1 \vdots x – 3\\
\to 2\left( {x – 3} \right) + 7 \vdots x – 3\\
\to 7 \vdots x – 3\\
\to x – 3 \in U\left( 7 \right)\\
\to \left[ \begin{array}{l}
x – 3 = 7\\
x – 3 = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 10\\
x = 4
\end{array} \right.\\
8)x – 1 \in U\left( {3x + 25} \right)\\
\to 3x + 25 \vdots x – 1\\
\to 3\left( {x – 1} \right) + 28 \vdots x – 1\\
\to 28 \vdots x – 1\\
\to x – 1 \in U\left( {28} \right)\\
\to \left[ \begin{array}{l}
x – 1 = 28\\
x – 1 = 14\\
x – 1 = 7\\
x – 1 = 4\\
x – 1 = 2\\
x – 1 = 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 29\\
x = 15\\
x = 8\\
x = 5\\
x = 3\\
x = 2
\end{array} \right.\\
9)2x + 3 \vdots 3x + 1\\
\to 6x + 9 \vdots 3x + 1\\
\to 2\left( {3x + 1} \right) + 7 \vdots 3x + 1\\
\to 7 \vdots 3x + 1\\
\to 3x + 1 \in U\left( 7 \right)\\
\to \left[ \begin{array}{l}
3x + 1 = 7\\
3x + 1 = 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 2\\
x = 0
\end{array} \right.\\
10)5x + 4 \in U\left( {4x – 2} \right)\\
\to 4x – 2 \vdots 5x + 4\\
\to 20x – 10 \vdots 5x + 4\\
\to 4\left( {5x + 4} \right) – 26 \vdots 5x + 4\\
\to 26 \vdots 5x + 4\\
\to \left[ \begin{array}{l}
5x + 4 = 26\\
5x + 4 = 13\\
5x + 4 = 2\\
5x + 4 = 1
\end{array} \right. \to \left[ \begin{array}{l}
x = \dfrac{{22}}{5}\\
x = \dfrac{9}{5}\\
x = – \dfrac{2}{5}\\
x = – \dfrac{3}{5}
\end{array} \right.
\end{array}\)