tìm số nguyên x, biết
1, (x + 3) chia hết ( x + 1 )
2, ( x ² – x + 2 ) chia hết ( x – 1 )
3, ( 2x + 5 ) chia hết ( x + 2 )
4, ( x ² + 2x + 4 ) chia hết ( x + 1 )
5, ( 3x + 5 ) chia hết ( x – 2 )
…..hết….
giúp em đi ạ
Đáp án:
5) \(\left[ \begin{array}{l}
x = 13\\
x = – 9\\
x = 3\\
x = 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1)x + 3 \vdots x + 1\\
\to x + 1 + 2 \vdots x + 1\\
\to 2 \vdots x + 1\\
\to x + 1 \in U\left( 2 \right)\\
\to \left[ \begin{array}{l}
x + 1 = 2\\
x + 1 = – 2\\
x + 1 = 1\\
x + 1 = – 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 1\\
x = – 3\\
x = 0\\
x = – 2
\end{array} \right.\\
2){x^2} – x + 2 \vdots x – 1\\
\to x\left( {x – 1} \right) + 2 \vdots x – 1\\
\to 2 \vdots x – 1\\
\to x – 1 \in U\left( 2 \right)\\
\to \left[ \begin{array}{l}
x – 1 = 2\\
x – 1 = – 2\\
x – 1 = 1\\
x – 1 = – 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 3\\
x = – 1\\
x = 2\\
x = 0
\end{array} \right.\\
3)2x + 5 \vdots x + 2\\
\to 2\left( {x + 2} \right) + 1 \vdots x + 2\\
\to 1 \vdots x + 2\\
\to x + 2 \in U\left( 1 \right)\\
\to \left[ \begin{array}{l}
x + 2 = 1\\
x + 2 = – 1
\end{array} \right. \to \left[ \begin{array}{l}
x = – 1\\
x = – 3
\end{array} \right.\\
4){x^2} + x + x + 4 \vdots x + 1\\
\to x\left( {x + 1} \right) + \left( {x + 1} \right) + 3 \vdots x + 1\\
\to 3 \vdots x + 1\\
\to x + 1 \in U\left( 3 \right)\\
\to \left[ \begin{array}{l}
x + 1 = 3\\
x + 1 = – 3\\
x + 1 = 1\\
x + 1 = – 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 2\\
x = – 4\\
x = 0\\
x = – 2
\end{array} \right.\\
5)3x + 5 \vdots x – 2\\
\to 3\left( {x – 2} \right) + 11 \vdots x – 2\\
\to 11 \vdots x – 2\\
\to x – 2 \in U\left( {11} \right)\\
\to \left[ \begin{array}{l}
x – 2 = 11\\
x – 2 = – 11\\
x – 2 = 1\\
x – 2 = – 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 13\\
x = – 9\\
x = 3\\
x = 1
\end{array} \right.
\end{array}\)