|x-1|=3x+2 |3x-1|+2=x |x+15|+1=3x |3x-2|-1=x |x+7|-x=7 20/07/2021 Bởi Peyton |x-1|=3x+2 |3x-1|+2=x |x+15|+1=3x |3x-2|-1=x |x+7|-x=7
Đáp án: a. \(\left[ \begin{array}{l}x = – \dfrac{3}{2}\\x = – \dfrac{1}{4}\end{array} \right.\) Giải thích các bước giải: \(\begin{array}{l}a.\left| {x – 1} \right| = 3x + 2\\ \to \left[ \begin{array}{l}x – 1 = 3x + 2\\x – 1 = – 3x – 2\end{array} \right.\\ \to \left[ \begin{array}{l}2x = – 3\\4x = – 1\end{array} \right.\\ \to \left[ \begin{array}{l}x = – \dfrac{3}{2}\\x = – \dfrac{1}{4}\end{array} \right.\\b.\left| {3x – 1} \right| = x – 2\\ \to \left[ \begin{array}{l}3x – 1 = x – 2\\3x – 1 = – x + 2\end{array} \right.\\ \to \left[ \begin{array}{l}2x = – 1\\4x = 3\end{array} \right.\\ \to \left[ \begin{array}{l}x = – \dfrac{1}{2}\\x = \dfrac{3}{4}\end{array} \right.\\c.\left| {x + 15} \right| = 3x – 1\\ \to \left[ \begin{array}{l}x + 15 = 3x – 1\\x + 15 = – 3x + 1\end{array} \right.\\ \to \left[ \begin{array}{l}2x = 16\\4x = – 14\end{array} \right.\\ \to \left[ \begin{array}{l}x = 8\\x = – \dfrac{7}{2}\end{array} \right.\\d.\left| {3x – 2} \right| = x + 1\\ \to \left[ \begin{array}{l}3x – 2 = x + 1\\3x – 2 = – x – 1\end{array} \right.\\ \to \left[ \begin{array}{l}2x = 3\\4x = 1\end{array} \right.\\ \to \left[ \begin{array}{l}x = \dfrac{3}{2}\\x = \dfrac{1}{4}\end{array} \right.\\e.\left| {x + 7} \right| = 7 + x\\ \to \left[ \begin{array}{l}x + 7 = x + 7\left( {ld} \right)\\x + 7 = – x – 7\end{array} \right.\\ \to 2x = – 14\\ \to x = – 7\end{array}\) Bình luận
Đáp án:
a. \(\left[ \begin{array}{l}
x = – \dfrac{3}{2}\\
x = – \dfrac{1}{4}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\left| {x – 1} \right| = 3x + 2\\
\to \left[ \begin{array}{l}
x – 1 = 3x + 2\\
x – 1 = – 3x – 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x = – 3\\
4x = – 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = – \dfrac{3}{2}\\
x = – \dfrac{1}{4}
\end{array} \right.\\
b.\left| {3x – 1} \right| = x – 2\\
\to \left[ \begin{array}{l}
3x – 1 = x – 2\\
3x – 1 = – x + 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x = – 1\\
4x = 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = – \dfrac{1}{2}\\
x = \dfrac{3}{4}
\end{array} \right.\\
c.\left| {x + 15} \right| = 3x – 1\\
\to \left[ \begin{array}{l}
x + 15 = 3x – 1\\
x + 15 = – 3x + 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x = 16\\
4x = – 14
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 8\\
x = – \dfrac{7}{2}
\end{array} \right.\\
d.\left| {3x – 2} \right| = x + 1\\
\to \left[ \begin{array}{l}
3x – 2 = x + 1\\
3x – 2 = – x – 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x = 3\\
4x = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{3}{2}\\
x = \dfrac{1}{4}
\end{array} \right.\\
e.\left| {x + 7} \right| = 7 + x\\
\to \left[ \begin{array}{l}
x + 7 = x + 7\left( {ld} \right)\\
x + 7 = – x – 7
\end{array} \right.\\
\to 2x = – 14\\
\to x = – 7
\end{array}\)