tìm các giá trị x để
1/ 1 chia hết cho x
2/ -2 chia hết cho x
3/ 1 chia hết cho (x+7)
4/ 4 chia hết cho (x-5)
5/ (x+7+1) chia hết cho ( x +7 )
6/ (x+8) chia hết cho (x+7)
7/ (2x+14+2) chia hết cho (x+7)
8/ (2x+16) chia hết cho (x+7)
9/ (x-5+1) chia hết cho (x-5)
10/ (2x-9) chia hết cho (x-5)
11/ (x^2 – x -1 ) chia hết cho (x-1)
12/3x là bội số của (x+1)
13/(x^2-3x-5)chia hết cho (x-3)
14/(5x+2)chia hết cho (x+1)
15/ (2x^2 +3x+2) chia hết cho (x+1)
các bn lm câu mk bt thoi cx dc nhanh lên nhé ( đầy đủ nha ) mk sẽ tick sao cho câu trả lời của các bn
Đáp án:
14) \(\left[ \begin{array}{l}
x = 2\\
x = 0
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1)1 \vdots x\\
\Leftrightarrow x \in U\left( 1 \right)\\
\to x = 1\\
2) – 2 \vdots x\\
\Leftrightarrow x \in U\left( 2 \right)\\
\to \left[ \begin{array}{l}
x = 2\\
x = 1
\end{array} \right.\\
3)1 \vdots x + 7\\
\to x + 7 \in U\left( 1 \right)\\
\to x + 7 = 1\\
\to x = – 6\\
4)4 \vdots x – 5\\
\to x – 5 \in U\left( 4 \right)\\
\to \left[ \begin{array}{l}
x – 5 = 4\\
x – 5 = 2\\
x – 5 = 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 9\\
x = 7\\
x = 6
\end{array} \right.\\
5)x + 7 + 1 \vdots x + 7\\
\to 1 \vdots x + 7\\
\to x + 7 \in U\left( 1 \right)\\
\to x + 7 = 1\\
\to x = – 6\\
6)x + 8 \vdots x + 7\\
\to x + 7 + 1 \vdots x + 7\\
\to 1 \vdots x + 7\\
\to x + 7 \in U\left( 1 \right)\\
\to x + 7 = 1\\
\to x = – 6\\
7)2x + 14 + 2 \vdots x + 7\\
\to 2\left( {x + 7} \right) + 2 \vdots x + 7\\
\to 2 \vdots x + 7\\
\to x + 7 \in U\left( 2 \right)\\
\to \left[ \begin{array}{l}
x + 7 = 2\\
x + 7 = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = – 5\\
x = – 6
\end{array} \right.\\
8)2x + 16 \vdots x + 7\\
\to 2x + 14 + 2 \vdots x + 7\\
\to 2\left( {x + 7} \right) + 2 \vdots x + 7\\
\to 2 \vdots x + 7\\
\to x + 7 \in U\left( 2 \right)\\
\to \left[ \begin{array}{l}
x + 7 = 2\\
x + 7 = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = – 5\\
x = – 6
\end{array} \right.\\
9)x – 5 + 1 \vdots x – 5\\
\to 1 \vdots x – 5\\
\to x – 5 \in U\left( 1 \right)\\
\to x – 5 = 1\\
\to x = 6\\
10)2x – 9 \vdots x – 5\\
\to 2\left( {x – 5} \right) + 1 \vdots x – 5\\
\to 1 \vdots x – 5\\
\to x – 5 \in U\left( 1 \right)\\
\to x – 5 = 1\\
\to x = 6\\
12)3x \in B\left( {x + 1} \right)\\
\to 3x \vdots x + 1\\
\to 3\left( {x + 1} \right) – 3 \vdots x + 1\\
\to 3 \vdots x + 1\\
\to x + 1 \in U\left( 3 \right)\\
\to \left[ \begin{array}{l}
x + 1 = 3\\
x + 1 = 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 2\\
x = 0
\end{array} \right.\\
14)5x + 2 \vdots x + 1\\
\to 5\left( {x + 1} \right) – 3 \vdots x + 1\\
\to 3 \vdots x + 1\\
\to x + 1 \in U\left( 3 \right)\\
\to \left[ \begin{array}{l}
x + 1 = 3\\
x + 1 = 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 2\\
x = 0
\end{array} \right.
\end{array}\)