Đáp án:

$\begin{array}{l}
a)3\left( {x - 2} \right) + 3 = x + 1\\
 \Rightarrow 3x - 6 + 3 = x + 1\\
 \Rightarrow 3x - x = 1 + 6 - 3\\
 \Rightarrow 2x = 4\\
 \Rightarrow x = 2\\
Vậy\,x = 2\\
b)\frac{{x + 1}}{2} = \frac{3}{{y + 2}}\\
 \Rightarrow \left( {x + 1} \right).\left( {y + 2} \right) = 3.2 = 1.6\\
 \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 1 = 3\\
y + 2 = 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 1 = 2\\
y + 2 = 3
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 1 = 1\\
y + 2 = 6
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 1 = 6\\
y + 2 = 1
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = 2;y = 0\\
x = 1;y = 1\\
x = 0;y = 4\\
x = 5;y =  - 1
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left\{ {\left( {2;0} \right);\left( {1;1} \right);\left( {0;4} \right);\left( {5; - 1} \right)} \right\}\\
c)xy - 3y + x = 3\\
 \Rightarrow y.\left( {x - 3} \right) + x - 3 = 0\\
 \Rightarrow \left( {x - 3} \right).\left( {y + 1} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
x = 3\\
y =  - 1
\end{array} \right.\\
Vậy\,x = 3;x =  - 1\\
d)\left| {x - 2} \right| + {\left| {y - 1} \right|^{100}} + {\left| {2z + 4} \right|^{209}} = 0\\
Do:\left| a \right| \ge 0\forall a\\
 \Rightarrow \left\{ \begin{array}{l}
\left| {x - 2} \right| = 0\\
\left| {y - 1} \right| = 0\\
\left| {2z + 4} \right| = 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
x = 2\\
y = 1\\
z =  - 2
\end{array} \right.\\
Vậy\,x = 2;y = 1;z =  - 2\\
e){\left( {x - 2} \right)^{2020}} + 3{\left| {xy + 4} \right|^{20}} = 0\\
 \Rightarrow \left\{ \begin{array}{l}
x - 2 = 0\\
xy + 4 = 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
x = 2\\
xy =  - 4
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
x = 2\\
y =  - 2
\end{array} \right.\\
Vậy\,x = 2;y =  - 2
\end{array}$