$$\begin{gathered} a, x-2\sqrt{x}+7=4, đk: x\ge 0 \\ <=>x-2\sqrt{x}-3=0 \\ <=>\left(x-3\sqrt{x}\right)+\left(\sqrt{x}-3\right)=0 \\ <=>\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}-3\right)=0 \\ <=>\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)=0 \\ <=>\sqrt{x}-3=0, do \sqrt{x}+1\ge 1>0 \\ <=>\sqrt{x}=3 \\ <=>x=9 \left(tm\right) \\ Vậy x=9 \\ b, \left|3x+1\right|=\left|2x-2\right| \\ <=>3x+1=2x-2 hoặc 3x+1=2-2x \\ <=>x=-3 hoặc 5x=1 \\ <=>x=-3 hoặc x=\frac{1}{5} \\ Vậy S=\left\{-3;\frac{1}{5}\right\} \\ c, \left|3x+2\right|+x=2 \\ +) 3x+2\ge 0<=>x\ge -\frac{2}{3} ta có: \\ 3x+2+x=2<=>4x=0<=>x=0 \left(tm\right) \\ +) 3x+2<0<=>x<-\frac{2}{3} ta có: \\ -3x-2+x=2<=>-2x=4<=>x=-2 \left(tm\right) \\ Vậy S=\left\{-2;0\right\} \end{gathered}$$..

\[\begin{array}{l}
A.{\rm{ }}x - \sqrt {2x + 7} = 4\\
dk:x \ge \frac{{ - 7}}{2}\\
\Leftrightarrow \sqrt {2{\rm{x}} + 7} = x - 4\\
\Leftrightarrow \left\{ \begin{array}{l}
x - 4 \ge 0\\
2{\rm{x}} + 7 = {\left( {x - 4} \right)^2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 4\\
2{\rm{x}} + 7 = {x^2} - 8{\rm{x}} + 16
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 4\\
{x^2} - 10{\rm{x}} + 9 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 4\\
\left( {x - 1} \right)\left( {x - 9} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 4\\
\left[ \begin{array}{l}
x - 1 = 0\\
x - 9 = 0
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 4\\
\left[ \begin{array}{l}
x = 1\\
x = 9
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow x = 9\left( {tm} \right)\\
vay:x = 9\\
B.{\rm{ }}\left| {3x + 1} \right| = \left| {2x - 2} \right|\\
\Leftrightarrow \left[ \begin{array}{l}
3{\rm{x}} + 1 = 2{\rm{x}} - 2\\
3{\rm{x}} + 1 = 2 - 2{\rm{x}}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
3{\rm{x}} - 2{\rm{x}} = - 2 - 1\\
3{\rm{x}} + 2{\rm{x}} = 2 - 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 3\\
x = \frac{1}{5}
\end{array} \right.\\
vay:S = \{ - 3;\frac{1}{5}\} \\
C.{\rm{ }}\left| {3x + 2} \right| + x = 2\\
\Leftrightarrow |3{\rm{x}} + 2| = 2 - x\\
\Leftrightarrow \left\{ \begin{array}{l}
2 - x \ge 0\\
\left[ \begin{array}{l}
3{\rm{x}} + 2 = 2 - x\\
3{\rm{x}} + 2 = x - 2
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \le 2\\
\left[ \begin{array}{l}
3{\rm{x}} + x = 2 - 2\\
{\rm{3x}} - x = - 2 - 2
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{x \le 2}\\
{\left[ \begin{array}{l}
4{\rm{x}} = 0\\
2x = - 4
\end{array} \right.}
\end{array}} \right.\\
\Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{x \le 2}\\
{\left[ \begin{array}{l}
x = 0\\
x = - 2
\end{array} \right.}
\end{array}} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = - 2
\end{array} \right.\\
vay:S = \{ - 2;0\}
\end{array}\]