A. \(x = 4\)
B. \(x = 3\)
C. \(x = 2\)
D. \(x = 5\)
B
\(\begin{array}{l}\,\,\left| {x - 1} \right| + 7 = 3x\\ \Leftrightarrow \left| {x - 1} \right| = 3x - 7\\ \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x - 1 \ge 0\\x - 1 = 3x - 7\end{array} \right.\\\left\{ \begin{array}{l}x - 1 < 0\\ - \left( {x - 1} \right) = 3x - 7\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x \ge 1\\x - 1 = 3x - 7\end{array} \right.\\\left\{ \begin{array}{l}x < 1\\ - x + 1 = 3x - 7\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x \ge 1\\3x - x = - 1 + 7\end{array} \right.\\\left\{ \begin{array}{l}x < 1\\x + 3x = 7 + 1\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x \ge 1\\2x = 6\end{array} \right.\\\left\{ \begin{array}{l}x < 1\\4x = 8\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x \ge 1\\x = 3\end{array} \right.\\\left\{ \begin{array}{l}x < 1\\x = 2\end{array} \right.\end{array} \right.\\ \Leftrightarrow x = 3.\end{array}\)
Vậy phương trình có nghiệm duy nhất \(x = 3.\)
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