Giải hệ phương trình:
$ \left\{ \begin{array}{l}
|x  -  y|  +  2|x  +  y  -  1|  =  3\\
2x  +  y  =  1
\end{array} \right.$                          
Từ phương trình thứ 2  $  \Rightarrow y = 1 - 2x $  thay vào phương trình đầu ta có: $ |3x - 1| + 2|x| = 3 $        (3)
Xét các khả năng sau :
1.    Nếu x < 0 $ \Rightarrow 1 - 3x - 2x = 3 $  $  \Leftrightarrow x = \frac{{ - 2}}{5} $ , thỏa mãn.
Nghiệm của hệ là : $ x =  - \frac{2}{5};y = \frac{9}{5} $
2.    Nếu  $ 0 \le x \le \frac{1}{3} $ : $ \Rightarrow 1 - 3x + 2x = 3 $  $  \Leftrightarrow x =  - 2 $ , không thỏa mãn.
3.    Nếu  $ x \ge \frac{1}{3} $  $ (3) \Rightarrow 3x - 1 + 2x = 3 $  $  \Leftrightarrow x = \frac{4}{5} $ , thỏa mãn.
Nghiệm của hệ là : $ x = \frac{4}{5};y =  - \frac{3}{5} $
Vậy hệ đã cho có 2 nghiệm là : $x=-\frac{2}{5};y = \frac{9}{5} $ và  $ x = \frac{4}{5};y =  - \frac{3}{5} $

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