Đáp án:

\(\begin{array}{l}
1,\\
\left[ \begin{array}{l}
x = 2\\
x = 0
\end{array} \right.\\
2,\\
\left[ \begin{array}{l}
x = 3\\
x = 1
\end{array} \right.\\
3,\\
\left[ \begin{array}{l}
x =  - 2\\
x = 3
\end{array} \right.\\
4,\\
\left[ \begin{array}{l}
x =  - 2\\
x = 4
\end{array} \right.\\
5,\\
\left[ \begin{array}{l}
x = 1\\
x =  - \dfrac{1}{5}
\end{array} \right.\\
6,\\
x = \dfrac{3}{2}
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
DKXD:\,\,\,{x^2} - 2x + 1 \ge 0 \Leftrightarrow {\left( {x - 1} \right)^2} \ge 0,\,\,\,\forall x\\
\sqrt {{x^2} - 2x + 1}  = 1\\
 \Leftrightarrow \sqrt {{x^2} - 2.x.1 + {1^2}}  = 1\\
 \Leftrightarrow \sqrt {{{\left( {x - 1} \right)}^2}}  = 1\\
 \Leftrightarrow \left| {x - 1} \right| = 1\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 1 = 1\\
x - 1 =  - 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = 0
\end{array} \right.\\
2,\\
DKXD:\,\,\,{x^2} - 4x + 4 \ge 0 \Leftrightarrow {\left( {x - 2} \right)^2} \ge 0,\,\,\,\forall x\\
\sqrt {{x^2} - 4x + 4}  = 1\\
 \Leftrightarrow \sqrt {{x^2} - 2.x.2 + {2^2}}  = 1\\
 \Leftrightarrow \sqrt {{{\left( {x - 2} \right)}^2}}  = 1\\
 \Leftrightarrow \left| {x - 2} \right| = 1\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 2 = 1\\
x - 2 =  - 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = 1
\end{array} \right.\\
3,\\
DKXD:\,\,\,1 - 4x + 4{x^2} \ge 0 \Leftrightarrow {\left( {1 - 2x} \right)^2} \ge 0,\,\,\,\forall x\\
\sqrt {1 - 4x + 4{x^2}}  = 5\\
 \Leftrightarrow \sqrt {{1^2} - 2.1.2x + {{\left( {2x} \right)}^2}}  = 5\\
 \Leftrightarrow \sqrt {{{\left( {1 - 2x} \right)}^2}}  = 5\\
 \Leftrightarrow \left| {1 - 2x} \right| = 5\\
 \Leftrightarrow \left[ \begin{array}{l}
1 - 2x = 5\\
1 - 2x =  - 5
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2x =  - 4\\
2x = 6
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x =  - 2\\
x = 3
\end{array} \right.\\
4,\\
DKXD:\,\,1 - 2x + {x^2} \ge 0 \Leftrightarrow {\left( {1 - x} \right)^2} \ge 0,\,\,\,\forall x\\
\sqrt {4\left( {1 - 2x + {x^2}} \right)}  - 6 = 0\\
 \Leftrightarrow \sqrt {{2^2}.\left( {{1^2} - 2.1.x + {x^2}} \right)}  - 6 = 0\\
 \Leftrightarrow \sqrt {{2^2}.{{\left( {1 - x} \right)}^2}}  = 6\\
 \Leftrightarrow 2.\left| {1 - x} \right| = 6\\
 \Leftrightarrow \left| {1 - x} \right| = 3\\
 \Leftrightarrow \left[ \begin{array}{l}
1 - x = 3\\
1 - x =  - 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x =  - 2\\
x = 4
\end{array} \right.\\
5,\\
DKXD:\,\,\,{x^2} \ge 0,\,\,\forall x\\
\sqrt {9{x^2}}  = 2x + 1\\
 \Leftrightarrow \sqrt {{{\left( {3x} \right)}^2}}  = 2x + 1\\
 \Leftrightarrow \left| {3x} \right| = 2x + 1\\
 \Leftrightarrow \left\{ \begin{array}{l}
2x + 1 \ge 0\\
\left[ \begin{array}{l}
3x = 2x + 1\\
3x =  - 2x - 1
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge  - \dfrac{1}{2}\\
\left[ \begin{array}{l}
3x - 2x = 1\\
3x + 2x =  - 1
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge  - \dfrac{1}{2}\\
\left[ \begin{array}{l}
x = 1\\
5x =  - 1
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge  - \dfrac{1}{2}\\
\left[ \begin{array}{l}
x = 1\\
x =  - \dfrac{1}{5}
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x =  - \dfrac{1}{5}
\end{array} \right.\\
6,\\
DKXD:\,\,\,9 - 6x + {x^2} \ge 0 \Leftrightarrow {\left( {3 - x} \right)^2} \ge 0,\,\,\,\forall x\\
\sqrt {9 - 6x + {x^2}}  = x\\
 \Leftrightarrow \sqrt {{3^2} - 2.3.x + {x^2}}  = x\\
 \Leftrightarrow \sqrt {{{\left( {3 - x} \right)}^2}}  = x\\
 \Leftrightarrow \left| {3 - x} \right| = x\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
\left\{ \begin{array}{l}
3 - x = x\\
3 - x =  - x
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
\left[ \begin{array}{l}
3 = x + x\\
3 = x - x
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
\left[ \begin{array}{l}
2x = 3\\
3 = 0\,\,\,\left( L \right)
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x = \dfrac{3}{2}
\end{array} \right. \Leftrightarrow x = \dfrac{3}{2}
\end{array}\)