Toán a) |x-5|=13-2x b)|5x-1|=x-12 c)|-2x|=3x+4 d)|2x-1|=6-x 09/10/2021 By Rylee a) |x-5|=13-2x b)|5x-1|=x-12 c)|-2x|=3x+4 d)|2x-1|=6-x
Đáp án: d. \(\left[ \begin{array}{l}x = \dfrac{7}{3}\\x = – 5\end{array} \right.\) Giải thích các bước giải: \(\begin{array}{l}a.\left| {x – 5} \right| = 13 – 2x\\ \to \left[ \begin{array}{l}x – 5 = 13 – 2x\left( {x \ge 5} \right)\\x – 5 = – 13 + 2x\left( {x < 5} \right)\end{array} \right.\\ \to \left[ \begin{array}{l}3x = 18\\x = 8\left( l \right)\end{array} \right.\\ \to x = 6\left( {TM} \right)\\b.\left| {5x – 1} \right| = x – 12\\ \to \left[ \begin{array}{l}5x – 1 = x – 12\left( {x \ge \dfrac{1}{5}} \right)\\5x – 1 = – x + 12\left( {x < \dfrac{1}{5}} \right)\end{array} \right.\\ \to \left[ \begin{array}{l}4x = – 11\\6x = 13\end{array} \right.\\ \to \left[ \begin{array}{l}x = – \dfrac{{11}}{4}\left( l \right)\\x = \dfrac{{13}}{6}\left( l \right)\end{array} \right.\end{array}\) ⇒ Phương trình vô nghiệm \(\begin{array}{l}c.\left| { – 2x} \right| = 3x + 4\\ \to \left[ \begin{array}{l} – 2x = 3x + 4\left( {x \ge 0} \right)\\2x = 3x + 4\left( {x < 0} \right)\end{array} \right.\\ \to \left[ \begin{array}{l}5x = – 4\\x = – 4\end{array} \right.\\ \to \left[ \begin{array}{l}x = – \dfrac{4}{5}\left( l \right)\\x = – 4\left( {TM} \right)\end{array} \right.\\d.\left| {2x – 1} \right| = 6 – x\\ \to \left[ \begin{array}{l}2x – 1 = 6 – x\left( {x \ge \dfrac{1}{2}} \right)\\2x – 1 = – 6 + x\left( {x < \dfrac{1}{2}} \right)\end{array} \right.\\ \to \left[ \begin{array}{l}3x = 7\\x = – 5\end{array} \right.\\ \to \left[ \begin{array}{l}x = \dfrac{7}{3}\\x = – 5\end{array} \right.\end{array}\) Trả lời
Đáp án:
Giải thích các bước giải:
Đáp án:
d. \(\left[ \begin{array}{l}
x = \dfrac{7}{3}\\
x = – 5
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\left| {x – 5} \right| = 13 – 2x\\
\to \left[ \begin{array}{l}
x – 5 = 13 – 2x\left( {x \ge 5} \right)\\
x – 5 = – 13 + 2x\left( {x < 5} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
3x = 18\\
x = 8\left( l \right)
\end{array} \right.\\
\to x = 6\left( {TM} \right)\\
b.\left| {5x – 1} \right| = x – 12\\
\to \left[ \begin{array}{l}
5x – 1 = x – 12\left( {x \ge \dfrac{1}{5}} \right)\\
5x – 1 = – x + 12\left( {x < \dfrac{1}{5}} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
4x = – 11\\
6x = 13
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = – \dfrac{{11}}{4}\left( l \right)\\
x = \dfrac{{13}}{6}\left( l \right)
\end{array} \right.
\end{array}\)
⇒ Phương trình vô nghiệm
\(\begin{array}{l}
c.\left| { – 2x} \right| = 3x + 4\\
\to \left[ \begin{array}{l}
– 2x = 3x + 4\left( {x \ge 0} \right)\\
2x = 3x + 4\left( {x < 0} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
5x = – 4\\
x = – 4
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = – \dfrac{4}{5}\left( l \right)\\
x = – 4\left( {TM} \right)
\end{array} \right.\\
d.\left| {2x – 1} \right| = 6 – x\\
\to \left[ \begin{array}{l}
2x – 1 = 6 – x\left( {x \ge \dfrac{1}{2}} \right)\\
2x – 1 = – 6 + x\left( {x < \dfrac{1}{2}} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
3x = 7\\
x = – 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{7}{3}\\
x = – 5
\end{array} \right.
\end{array}\)