Đáp án:
$\begin{array}{l}
e)Dkxd:x \ge 0\\
\left| {x - 2} \right| = x\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2 = x\left( {khi:x \ge 2} \right)\\
2 - x = x\left( {khi:0 \le x < 2} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
- 2 = 0\left( {ktm} \right)\\
2x = 2 \Leftrightarrow x = 1\left( {tmdk} \right)
\end{array} \right.\\
Vậy\,x = 1\\
f)Dkxd:x \ge 0\\
\left| {x - 3,4} \right| + \left| {2,6 - x} \right| = x\\
\Leftrightarrow \left| {x - 3,4} \right| + \left| {x - 2,6} \right| = x\\
+ Khi:x \ge 3,4\\
\Leftrightarrow x - 3,4 + x - 2,6 = x\\
\Leftrightarrow x = 6\left( {tmdk} \right)\\
+ Khi:2,6 \le x < 3,4\\
\Leftrightarrow 3,4 - x + x - 2,6 = x\\
\Leftrightarrow x = 0,8\left( {ktm} \right)\\
+ Khi:0 \le x < 2,6\\
\Leftrightarrow 3,4 - x + 2,6 - x = x\\
\Leftrightarrow 3x = 6\\
\Leftrightarrow x = 2\left( {tmdk} \right)\\
Vậy\,x = 2;x = 6
\end{array}$
