Đáp án: 10.D 11.B 12.C
Giải thích các bước giải:
$\begin{array}{l}
10)Sai\\
Do:\left\{ \begin{array}{l}
\left| {2x - 4} \right| \ge 0\forall x\\
\left| {x - 1} \right| \ge 0\forall x
\end{array} \right.\\
\left| {2x - 4} \right| + \left| {x - 1} \right| = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
2x - 4 = 0\\
x - 1 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 2\\
x = 1
\end{array} \right.\\
\Rightarrow x \in \emptyset \\
\Rightarrow D\\
11)\,Đúng.\\
12)Sai\\
4x\left( {x - 1} \right) = \left| {2x - 1} \right| + 1\\
\Leftrightarrow \left[ \begin{array}{l}
4{x^2} - 4x = 2x - 1 + 1\left( {khi\,x \ge \frac{1}{2}} \right)\\
4{x^2} - 4x = 1 - 2x + 1\left( {khi\,x < \frac{1}{2}} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
4{x^2} - 6x = 0\left( {khi\,x \ge \frac{1}{2}} \right)\\
4{x^2} - 2x - 2 = 0\left( {khi\,x < \frac{1}{2}} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{3}{2}\\
x = \frac{{ - 1}}{2}
\end{array} \right.\\
\Rightarrow S = \frac{3}{2} - \frac{1}{2} = 1\\
\Rightarrow C
\end{array}$
