Đáp án:
$\begin{array}{l}
a)\left| {3x - 5} \right| + {\left( {2y + 5} \right)^{208}} + {\left( {4z - 3} \right)^{20}} \le 0\\
Do:\left\{ \begin{array}{l}
\left| {3x - 5} \right| \ge 0\\
{\left( {2y + 5} \right)^{208}} \ge 0\\
{\left( {4z - 3} \right)^{20}} \ge 0
\end{array} \right.\\
Khi:\left| {3x - 5} \right| + {\left( {2y + 5} \right)^{208}} + {\left( {4z - 3} \right)^{20}} \le 0\\
\Leftrightarrow \left\{ \begin{array}{l}
\left| {3x - 5} \right| = 0\\
{\left( {2y + 5} \right)^{208}} = 0\\
{\left( {4z - 3} \right)^{20}} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{5}{3}\\
y = - \dfrac{5}{2}\\
z = \dfrac{3}{4}
\end{array} \right.\\
Vậy\,x = \dfrac{5}{3};y = \dfrac{{ - 5}}{2};z = \dfrac{3}{4}\\
b)\dfrac{{45 - x}}{{1963}} + \dfrac{{40 - x}}{{1968}} + \dfrac{{35 - x}}{{1973}} + \dfrac{{30 - x}}{{1978}} + 4 = 0\\
\Leftrightarrow \dfrac{{45 - x}}{{1963}} + 1 + \dfrac{{40 - x}}{{1968}} + 1 + \dfrac{{35 - x}}{{1973}} + 1 + \dfrac{{30 - x}}{{1978}} + 1 = 0\\
\Leftrightarrow \dfrac{{2008 - x}}{{1963}} + \dfrac{{2008 - x}}{{1968}} + \dfrac{{2008 - x}}{{1973}} + \dfrac{{2008 - x}}{{1978}} = 0\\
\Leftrightarrow \left( {2008 - x} \right)\left( {\dfrac{1}{{1963}} + \dfrac{1}{{1968}} + \dfrac{1}{{1973}} + \dfrac{1}{{1978}}} \right) = 0\\
\Leftrightarrow 2008 - x = 0\\
\Leftrightarrow x = 2008\\
Vậy\,x = 2008
\end{array}$
