a) \(\left|3x-1\right|=\left|x+3\right|\)
⇒ \(\left[{}\begin{matrix}3x-1=x+3\\3x-1=-\left(x+3\right)\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}3x-x=3+1\\3x-1=-x-3\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}2x=4\\3x+x=\left(-3\right)+1\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=4:2\\4x=-2\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=2\\x=\left(-2\right):4\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x=2\\x=-\frac{1}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{2;-\frac{1}{2}\right\}.\)
b) \(\left|x-1\right|+3x=1\)
⇒ \(\left|x-1\right|=1-3x\)
⇒ \(\left[{}\begin{matrix}x-1=1-3x\\x-1=3x-1\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x+3x=1+1\\x-3x=\left(-1\right)+1\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}4x=2\\-2x=0\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x=2:4\\x=0:\left(-2\right)\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=\frac{1}{2}\\x=0\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{2};0\right\}.\)
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c, \(\left|5x-3\right|-x=7\)
\(\Rightarrow\left[{}\begin{matrix}5x-3-x=7\\5x-3-x=-7\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}4x-3=7\\4x-3=-7\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}4x=7+3\\4x=-7+3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}4x=10\\4x=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=10:4\\x=-4:4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=-1\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{5}{2};-1\right\}\)
