Giải thích các bước giải:

 b) Ta có:

$\begin{array}{l}
\dfrac{{x + 4}}{{2000}} + \dfrac{{x + 3}}{{2001}} + \dfrac{{x + 2}}{{2002}} = \dfrac{{x + 1}}{{2003}} + \dfrac{x}{{2004}} + 1\\
 \Leftrightarrow \dfrac{{x + 2004}}{{2000}} + \dfrac{{x + 2004}}{{2001}} + \dfrac{{x + 2004}}{{2002}} = \dfrac{{x + 1}}{{2003}} + \dfrac{x}{{2004}} + 3\\
 \Leftrightarrow \dfrac{{x + 2004}}{{2000}} + \dfrac{{x + 2004}}{{2001}} + \dfrac{{x + 2004}}{{2002}} = \dfrac{{x + 2004}}{{2003}} + \dfrac{{x + 2004}}{{2004}}+1(1)
\end{array}$

+)TH1: Nếu $x+2004=0$

Thì $(1)$ trở thành: $0=1(mt)$

$\to x=-2004$ loại.

+)TH2: Nếu $x+2004\ne 0$

Khi đó:

$\begin{array}{l}
\left( 1 \right) \Leftrightarrow \dfrac{{x + 2004}}{{2000}} + \dfrac{{x + 2004}}{{2001}} + \dfrac{{x + 2004}}{{2002}} = \dfrac{{x + 2004}}{{2003}} + \dfrac{{x + 2004}}{{2004}} + \dfrac{{x + 2004}}{{x + 2004}}\\
 \Leftrightarrow \dfrac{1}{{2000}} + \dfrac{1}{{2001}} + \dfrac{1}{{2002}} = \dfrac{1}{{2003}} + \dfrac{1}{{2004}} + \dfrac{1}{{x + 2004}}\\
 \Leftrightarrow \dfrac{1}{{x + 2004}} = \dfrac{1}{{2000}} + \dfrac{1}{{2001}} + \dfrac{1}{{2002}} - \left( {\dfrac{1}{{2003}} + \dfrac{1}{{2004}}} \right)\\
 \Leftrightarrow x + 2004 = \dfrac{1}{{\dfrac{1}{{2000}} + \dfrac{1}{{2001}} + \dfrac{1}{{2002}} - \dfrac{1}{{2003}} - \dfrac{1}{{2004}}}}\\
 \Leftrightarrow x + 2004 = 1996\\
 \Leftrightarrow x =  - 8
\end{array}$

Vậy $x=-8$

c) Ta có:

${\left| {2x - \dfrac{1}{3}} \right|^{2021}} + 2021{\left( {x - y} \right)^{2020}} = 0\left( 1 \right)$

Lại có:

$\begin{array}{l}
\left\{ \begin{array}{l}
\left| {2x - \dfrac{1}{3}} \right| \ge 0 \Rightarrow {\left| {2x - \dfrac{1}{3}} \right|^{2021}} \ge 0\\
{\left( {x - y} \right)^{2020}} \ge 0 \Rightarrow 2021{\left( {x - y} \right)^{2020}} \ge 0
\end{array} \right.\\
 \Rightarrow {\left| {2x - \dfrac{1}{3}} \right|^{2021}} + 2021{\left( {x - y} \right)^{2020}} \ge 0
\end{array}$

Như vậy:

$\begin{array}{l}
\left( 1 \right) \Leftrightarrow {\left| {2x - \dfrac{1}{3}} \right|^{2021}} = 2021{\left( {x - y} \right)^{2020}} = 0\\
 \Leftrightarrow 2x - \dfrac{1}{3} = x - y = 0\\
 \Leftrightarrow x = y = \dfrac{1}{6}
\end{array}$

Vậy $\left( {x;y} \right) = \left( {\dfrac{1}{6};\dfrac{1}{6}} \right)$

d) Ta có:

${\left( {4x} \right)^{2020}} + 2022{\left( {x - 2y} \right)^{2020}} \le 0\left( 1 \right)$

Lại có:

$\begin{array}{l}
\left\{ \begin{array}{l}
{\left( {4x} \right)^{2020}} \ge 0\\
{\left( {x - 2y} \right)^{2020}} \ge 0
\end{array} \right.\\
 \Rightarrow {\left( {4x} \right)^{2020}} + 2022{\left( {x - 2y} \right)^{2020}} \ge 0
\end{array}$

Như vậy:

$\begin{array}{l}
\left( 1 \right) \Leftrightarrow {\left( {4x} \right)^{2020}} = 2022{\left( {x - 2y} \right)^{2020}} = 0\\
 \Leftrightarrow x = x - 2y = 0\\
 \Leftrightarrow x = y = 0
\end{array}$

Vậy $\left( {x;y} \right) = \left( {0;0} \right)$