Đáp án:
$\begin{array}{l}
2)Dkxd:4{x^2} - 4x + 1 \ge 0\\
\Leftrightarrow {\left( {2x - 1} \right)^2} \ge 0\left( {tm} \right)\\
\sqrt {4{x^2} - 4x + 1} - 5 = 0\\
\Leftrightarrow \sqrt {{{\left( {2x - 1} \right)}^2}} = 5\\
\Leftrightarrow \left| {2x - 1} \right| = 5\\
\Leftrightarrow \left[ \begin{array}{l}
2x - 1 = 5\\
2x - 1 = - 5
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = 2
\end{array} \right.\\
Vậy\,x = 2;x = 3\\
3)Dkxd:\left\{ \begin{array}{l}
{x^2} - 2x + 1 \ge 0\left( {tm} \right)\\
x - 1 \ge 0
\end{array} \right.\\
\Leftrightarrow x \ge 1\\
\sqrt {{x^2} - 2x + 1} = x - 1\\
\Leftrightarrow \sqrt {{{\left( {x - 1} \right)}^2}} = x - 1\\
\Leftrightarrow \left| {x - 1} \right| = x - 1\\
Vậy\,x \ge 1\\
4)\sqrt {16 - 8x + {x^2}} + x = 4\\
\Leftrightarrow \sqrt {{{\left( {x - 4} \right)}^2}} = 4 - x\\
\Leftrightarrow \left| {x - 4} \right| = 4 - x\\
\Leftrightarrow x - 4 \le 0\\
\Leftrightarrow x \le 4\\
Vậy\,x \le 4\\
5)\sqrt {{x^2} + x + \dfrac{1}{4}} = - x - \dfrac{1}{2}\\
\Leftrightarrow \sqrt {{{\left( {x + \dfrac{1}{2}} \right)}^2}} = - \left( {x + \dfrac{1}{2}} \right)\\
\Leftrightarrow x + \dfrac{1}{2} \le 0\\
\Leftrightarrow x \le - \dfrac{1}{2}\\
Vậy\,x \le - \dfrac{1}{2}
\end{array}$
