/7+3x\=/-2+4x\ /3x-7\=6x-1 /9-2x\=3+5x /4x+11\=12x-4 18/07/2021 Bởi Parker /7+3x\=/-2+4x\ /3x-7\=6x-1 /9-2x\=3+5x /4x+11\=12x-4
Đáp án: c. \(\left[ \begin{array}{l}x = \dfrac{6}{7}\\x = – 4\end{array} \right.\) Giải thích các bước giải: \(\begin{array}{l}a.\left| {3x + 7} \right| = \left| { – 2 + 4x} \right|\\ \to \left[ \begin{array}{l}3x + 7 = – 2 + 4x\\3x + 7 = 2 – 4x\end{array} \right.\\ \to \left[ \begin{array}{l}x = 9\\7x = – 5\end{array} \right.\\ \to \left[ \begin{array}{l}x = 9\\x = – \dfrac{5}{7}\end{array} \right.\\b.\left| {3x – 7} \right| = 6x – 1\\ \to \left[ \begin{array}{l}3x – 7 = 6x – 1\\3x – 7 = – 6x + 1\end{array} \right.\\ \to \left[ \begin{array}{l}3x = – 6\\9x = 8\end{array} \right.\\ \to \left[ \begin{array}{l}x = – 2\\x = \dfrac{8}{9}\end{array} \right.\\c.\left| {9 – 2x} \right| = 3 + 5x\\ \to \left[ \begin{array}{l}9 – 2x = 3 + 5x\\9 – 2x = – 3 – 5x\end{array} \right.\\ \to \left[ \begin{array}{l}7x = 6\\3x = – 12\end{array} \right.\\ \to \left[ \begin{array}{l}x = \dfrac{6}{7}\\x = – 4\end{array} \right.\\d.\left| {4x + 11} \right| = 12x – 4\\ \to \left[ \begin{array}{l}4x + 11 = 12x – 4\\4x + 11 = – 12x + 4\end{array} \right.\\ \to \left[ \begin{array}{l}8x = 15\\16x = – 7\end{array} \right.\\ \to \left[ \begin{array}{l}x = \dfrac{{15}}{8}\\x = – \dfrac{7}{{16}}\end{array} \right.\end{array}\) Bình luận
Đáp án:
c. \(\left[ \begin{array}{l}
x = \dfrac{6}{7}\\
x = – 4
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\left| {3x + 7} \right| = \left| { – 2 + 4x} \right|\\
\to \left[ \begin{array}{l}
3x + 7 = – 2 + 4x\\
3x + 7 = 2 – 4x
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 9\\
7x = – 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 9\\
x = – \dfrac{5}{7}
\end{array} \right.\\
b.\left| {3x – 7} \right| = 6x – 1\\
\to \left[ \begin{array}{l}
3x – 7 = 6x – 1\\
3x – 7 = – 6x + 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
3x = – 6\\
9x = 8
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = – 2\\
x = \dfrac{8}{9}
\end{array} \right.\\
c.\left| {9 – 2x} \right| = 3 + 5x\\
\to \left[ \begin{array}{l}
9 – 2x = 3 + 5x\\
9 – 2x = – 3 – 5x
\end{array} \right.\\
\to \left[ \begin{array}{l}
7x = 6\\
3x = – 12
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{6}{7}\\
x = – 4
\end{array} \right.\\
d.\left| {4x + 11} \right| = 12x – 4\\
\to \left[ \begin{array}{l}
4x + 11 = 12x – 4\\
4x + 11 = – 12x + 4
\end{array} \right.\\
\to \left[ \begin{array}{l}
8x = 15\\
16x = – 7
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{15}}{8}\\
x = – \dfrac{7}{{16}}
\end{array} \right.
\end{array}\)