\(\left|3x-1\right|=2x-3\\ \Rightarrow\left[{}\begin{matrix}3x-1=2x-3\\3x-1=3-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}3x-2x=1-3\\3x+2x=1+3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-2\\5x=4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{4}{5}\end{matrix}\right.\)
\(\left|3x-1\right|=2x-3\left(\text{đ}k:2x-3\ge0\Rightarrow x\ge\dfrac{3}{2}\right)\\\Rightarrow\left[{}\begin{matrix}3x-1=2x-3\\3x-1=3-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}3x-2x=1-3\\3x+2x=1+3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-2\\5x=4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2(L)\\x=\dfrac{4}{5}(L)\end{matrix}\right.\)
Free time - Answer #6:
`| 3x - 1 | = 2x - 3 (`Điều kiện: `2x - 3 >= 0 <=> x >= 3/2``)`
<=> \(\left[{}\begin{matrix}3x-1=2x-3\\3x-1=-2x+3\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}3x-2x=1-3\\3x+2x=1+3\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=-2\\5x=4\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=-2\left(Loai\right)\\x=\dfrac{4}{5}\left(Loai\right)\end{matrix}\right.\)
Vậy không tồn tại `x`
