Giải phương trình sau a, l-8x l=3x+2 b, l x-5 l=5x-1 c, l 6x l=2x+12 d, l x-4 l=3x-5 26/10/2021 Bởi Parker Giải phương trình sau a, l-8x l=3x+2 b, l x-5 l=5x-1 c, l 6x l=2x+12 d, l x-4 l=3x-5
Đáp án: d. \(\left[ \begin{array}{l}x = – \frac{1}{2}\\x = \frac{4}{9}\end{array} \right.\) Giải thích các bước giải: \(\begin{array}{l}a.\left| {8x} \right| = 3x + 2\\ \to \left[ \begin{array}{l}8x = 3x + 2\\8x = – 3x – 2\end{array} \right.\\ \to \left[ \begin{array}{l}5x = 2\\11x = – 2\end{array} \right.\\ \to \left[ \begin{array}{l}x = \frac{2}{5}\\x = – \frac{2}{{11}}\end{array} \right.\\b.\left| {x – 5} \right| = 5x – 1\\ \to \left[ \begin{array}{l}x – 5 = 5x – 1\\x – 5 = – 5x + 1\end{array} \right.\\ \to \left[ \begin{array}{l}4x = – 4\\6x = 6\end{array} \right.\\ \to \left[ \begin{array}{l}x = – 1\\x = 1\end{array} \right.\\c.\left| {6x} \right| = 2x + 12\\ \to \left[ \begin{array}{l}6x = 2x + 12\\6x = – 2x – 12\end{array} \right.\\ \to \left[ \begin{array}{l}4x = 12\\8x = – 12\end{array} \right.\\ \to \left[ \begin{array}{l}x = 3\\x = – \frac{3}{2}\end{array} \right.\\d.\left| {x – 4} \right| = 3x – 5\\ \to \left[ \begin{array}{l}x – 4 = 3x – 5\\x – 4 = – 3x + 5\end{array} \right.\\ \to \left[ \begin{array}{l}2x = – 1\\4x = 9\end{array} \right.\\ \to \left[ \begin{array}{l}x = – \frac{1}{2}\\x = \frac{4}{9}\end{array} \right.\end{array}\) Bình luận
Đáp án:
Giải thích các bước giải:
Đáp án:
d. \(\left[ \begin{array}{l}
x = – \frac{1}{2}\\
x = \frac{4}{9}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\left| {8x} \right| = 3x + 2\\
\to \left[ \begin{array}{l}
8x = 3x + 2\\
8x = – 3x – 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
5x = 2\\
11x = – 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \frac{2}{5}\\
x = – \frac{2}{{11}}
\end{array} \right.\\
b.\left| {x – 5} \right| = 5x – 1\\
\to \left[ \begin{array}{l}
x – 5 = 5x – 1\\
x – 5 = – 5x + 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
4x = – 4\\
6x = 6
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = – 1\\
x = 1
\end{array} \right.\\
c.\left| {6x} \right| = 2x + 12\\
\to \left[ \begin{array}{l}
6x = 2x + 12\\
6x = – 2x – 12
\end{array} \right.\\
\to \left[ \begin{array}{l}
4x = 12\\
8x = – 12
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
x = – \frac{3}{2}
\end{array} \right.\\
d.\left| {x – 4} \right| = 3x – 5\\
\to \left[ \begin{array}{l}
x – 4 = 3x – 5\\
x – 4 = – 3x + 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x = – 1\\
4x = 9
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = – \frac{1}{2}\\
x = \frac{4}{9}
\end{array} \right.
\end{array}\)