@Thỏ
Ta có:
I`x` - `3`I = I`3x` - `1`I
⇒\(\left[ \begin{array}{l}x - 3 = 3x - 1\\x - 3 = -3x + 1\end{array} \right.\)
⇒ \(\left[ \begin{array}{l}x - 3x = -1 + 3\\x + 3x = 1 + 3\end{array} \right.\)
⇒ \(\left[ \begin{array}{l}-2x = 2\\4x = 4\end{array} \right.\)
⇒ \(\left[ \begin{array}{l}x=-1\\x=1\end{array} \right.\)
Vậy `x` `∈` `{-1 ; 1}`
