Giải bất phương trình  $ |x - 2| \le |x + 4| $ 
Xét các  khả năng
1. Nếu  $ x < - 4 $ :

Ta có bpt tương đương:  $ 2 - x \le -(x + 4) $  $  \Leftrightarrow 0x \le - 6$: loại.

2. Nếu  $  - 4 \le x < 2 $
Ta có :  $ 2 - x \le x + 4 $  $  \Leftrightarrow x \ge  - 1 $  $  \Rightarrow  $ nghiệm phải tìm là : $  - 1 \le x < 2 $      (a)

3.    Nếu  $ x \ge 2 $

Ta có : $ x - 2 \le x + 4 $  $  \Leftrightarrow 0x < 6 $, thỏa mãn.
 $  \Rightarrow  $ nghiệm phải tìm là: $ x \ge 2 $                 (b)
Vậy nghiệm của bất phương trình đã cho là: $\begin{array}{l}- 1 \le x < 2 \vee x \ge 2
 \Leftrightarrow x \ge  - 1
\end{array} $

Cách khác:
Bình phương 2 vế ta có: $ \begin{array}{l}
|x - 2| \le |x + 4|
 \Leftrightarrow {x^2} - 4x + 4 \le {x^2} + 8x + 16
 \Leftrightarrow x \ge  - 1
\end{array} $

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