Đáp án:

$\begin{array}{l}
a)\dfrac{x}{7} = \dfrac{3}{x}\\
 \Rightarrow x.x = 3.7\\
 \Rightarrow {x^2} = 21\\
Vậy\,{x^2} = 21\\
b)\dfrac{{x + 5}}{3} = \dfrac{5}{9}\\
 \Rightarrow 9.\left( {x + 5} \right) = 3.5\\
 \Rightarrow x + 5 = \dfrac{5}{3}\\
 \Rightarrow x = \dfrac{5}{3} - 5\\
 \Rightarrow x = \dfrac{{ - 10}}{3}\\
Vậy\,x = \dfrac{{ - 10}}{3}\\
c)\dfrac{1}{x} + \dfrac{y}{2} = \dfrac{5}{8}\\
 \Rightarrow \dfrac{{2 + x.y}}{{2.x}} = \dfrac{5}{8}\\
 \Rightarrow 16 + 8.x.y = 10x\\
 \Rightarrow 8xy - 10x =  - 16\\
 \Rightarrow 2x.\left( {4y - 5} \right) =  - 8\\
 \Rightarrow x.\left( {4y - 5} \right) =  - 4\\
 \Rightarrow x.\left( {5 - 4y} \right) = 4 = 1.4 = 2.2
\end{array}$

$\begin{array}{l}
 \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = 4\\
5 - 4y = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x =  - 4\\
5 - 4y =  - 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 1\\
5 - 4y = 4
\end{array} \right.\\
\left\{ \begin{array}{l}
x =  - 1\\
5 - 4y =  - 4
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 2\\
5 - 4y = 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x =  - 2\\
5 - 4y =  - 2
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = 4\\
4y = 4
\end{array} \right.\\
\left\{ \begin{array}{l}
x =  - 4\\
4y = 6
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 1\\
4y = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x =  - 1\\
4y = 9
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 2\\
4y = 3
\end{array} \right.\\
\left\{ \begin{array}{l}
x =  - 2\\
4y = 7
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = 4\\
y = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x =  - 4\\
y = \dfrac{3}{2}
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 1\\
y = \dfrac{1}{4}
\end{array} \right.\\
\left\{ \begin{array}{l}
x =  - 1\\
y = \dfrac{9}{4}
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 2\\
y = \dfrac{3}{4}
\end{array} \right.\\
\left\{ \begin{array}{l}
x =  - 2\\
y = \dfrac{7}{4}
\end{array} \right.
\end{array} \right.\\
Vậy\left( {x;y} \right) = \left\{ \begin{array}{l}
\left( {4;1} \right);\left( { - 4;\dfrac{3}{2}} \right);\left( {1;\dfrac{1}{4}} \right);\\
\left( { - 1;\dfrac{9}{4}} \right);\left( {2;\dfrac{3}{4}} \right);\left( { - 2;\dfrac{7}{4}} \right)
\end{array} \right\}
\end{array}$

$\begin{array}{l}
d)\left| {2x - 1} \right| + 5 = 8\\
 \Rightarrow \left| {2x - 1} \right| = 3\\
 \Rightarrow \left[ \begin{array}{l}
2x - 1 = 3\\
2x - 1 =  - 3
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
2x = 4\\
2x =  - 2
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = 2\\
x =  - 1
\end{array} \right.\\
Vậy\,x = 2;x =  - 1
\end{array}$