Tìm x:a) 3 x -4 là bội của 13 – x b) 45-4 x chia hết cho 5 – 6x 19/11/2021 Bởi Rose Tìm x:a) 3 x -4 là bội của 13 – x b) 45-4 x chia hết cho 5 – 6x
Đáp án: a) \(\left[ \begin{array}{l}x = – 22\\x = 48\\x = 6\\x = 20\\x = 8\\x = 18\\x = 12\\x = 14\end{array} \right.\) Giải thích các bước giải: a) Do 3x-4 là bội của 13-x \(\begin{array}{l} \to 3x – 4 \vdots 13 – x\\ \to – 3\left( { – x + 13} \right) + 35 \vdots 13 – x\\ \to 35 \vdots 13 – x\\ \to 13 – x \in U\left( {35} \right)\\ \to \left[ \begin{array}{l}13 – x = 35\\13 – x = – 35\\13 – x = 7\\13 – x = – 7\\13 – x = 5\\13 – x = – 5\\13 – x = 1\\13 – x = – 1\end{array} \right. \to \left[ \begin{array}{l}x = – 22\\x = 48\\x = 6\\x = 20\\x = 8\\x = 18\\x = 12\\x = 14\end{array} \right.\\b)45 – 4x \vdots 5 – 6x\\ \to 135 – 12x \vdots 5 – 6x\\ \to 2\left( {5 – 6x} \right) + 125 \vdots 5 – 6x\\ \to 125 \vdots 5 – 6x\\ \to 5 – 6x \in U\left( {125} \right)\\ \to \left[ \begin{array}{l}5 – 6x = 125\\5 – 6x = 25\\5 – 6x = 5\\5 – 6x = 1\\5 – 6x = – 1\\5 – 6x = – 5\\5 – 6x = – 25\\5 – 6x = – 125\end{array} \right. \to \left[ \begin{array}{l}x = – 20\\x = – \dfrac{{10}}{3}\\x = 0\\x = \dfrac{2}{3}\\x = 1\\x = \dfrac{5}{3}\\x = 5\\x = \dfrac{{65}}{3}\end{array} \right.\end{array}\) Bình luận
Đáp án:
a) \(\left[ \begin{array}{l}
x = – 22\\
x = 48\\
x = 6\\
x = 20\\
x = 8\\
x = 18\\
x = 12\\
x = 14
\end{array} \right.\)
Giải thích các bước giải:
a) Do 3x-4 là bội của 13-x
\(\begin{array}{l}
\to 3x – 4 \vdots 13 – x\\
\to – 3\left( { – x + 13} \right) + 35 \vdots 13 – x\\
\to 35 \vdots 13 – x\\
\to 13 – x \in U\left( {35} \right)\\
\to \left[ \begin{array}{l}
13 – x = 35\\
13 – x = – 35\\
13 – x = 7\\
13 – x = – 7\\
13 – x = 5\\
13 – x = – 5\\
13 – x = 1\\
13 – x = – 1
\end{array} \right. \to \left[ \begin{array}{l}
x = – 22\\
x = 48\\
x = 6\\
x = 20\\
x = 8\\
x = 18\\
x = 12\\
x = 14
\end{array} \right.\\
b)45 – 4x \vdots 5 – 6x\\
\to 135 – 12x \vdots 5 – 6x\\
\to 2\left( {5 – 6x} \right) + 125 \vdots 5 – 6x\\
\to 125 \vdots 5 – 6x\\
\to 5 – 6x \in U\left( {125} \right)\\
\to \left[ \begin{array}{l}
5 – 6x = 125\\
5 – 6x = 25\\
5 – 6x = 5\\
5 – 6x = 1\\
5 – 6x = – 1\\
5 – 6x = – 5\\
5 – 6x = – 25\\
5 – 6x = – 125
\end{array} \right. \to \left[ \begin{array}{l}
x = – 20\\
x = – \dfrac{{10}}{3}\\
x = 0\\
x = \dfrac{2}{3}\\
x = 1\\
x = \dfrac{5}{3}\\
x = 5\\
x = \dfrac{{65}}{3}
\end{array} \right.
\end{array}\)