Giải thích các bước giải:

\(\begin{array}{l}
d,\\
\frac{7}{{2x - 3}} + \frac{1}{{2x - 2}} = \frac{3}{{x - 1}}\,\,\,\,\,\,\,\,\left( {x \ne \frac{3}{2};\,\,x \ne 1} \right)\\
 \Leftrightarrow \frac{7}{{2x - 3}} = \frac{3}{{x - 1}} - \frac{1}{{2x - 2}}\\
 \Leftrightarrow \frac{7}{{2x - 3}} = \frac{1}{{x - 1}}\left( {3 - \frac{1}{2}} \right)\\
 \Leftrightarrow \frac{7}{{2x - 3}} = \frac{5}{{2\left( {x - 1} \right)}}\\
 \Leftrightarrow 14\left( {x - 1} \right) = 5\left( {2x - 3} \right)\\
 \Leftrightarrow x =  - \frac{1}{4}\\
e,\\
7 + \frac{{5x}}{3} = x - 2\\
 \Leftrightarrow \frac{5}{3}x - x =  - 2 - 7\\
 \Leftrightarrow \frac{2}{3}x =  - 9\\
 \Leftrightarrow x =  - \frac{{27}}{2}\\
f,\\
\frac{{2x}}{3} - \frac{{2x + 5}}{6} = \frac{1}{2}\\
 \Leftrightarrow \frac{{2.2x - \left( {2x + 5} \right)}}{6} = \frac{1}{2}\\
 \Leftrightarrow 2x - 5 = 3\\
 \Leftrightarrow x = 4\\
g,\\
\frac{4}{{2x - 3}} - \frac{7}{{3x - 5}} = 0\,\,\,\,\,\,\,\,\,\,\left( {x \ne \frac{3}{2};\,\,\,x \ne \frac{5}{3}} \right)\\
 \Leftrightarrow \frac{4}{{2x - 3}} = \frac{7}{{3x - 5}}\\
 \Leftrightarrow 4\left( {3x - 5} \right) = 7\left( {2x - 3} \right)\\
 \Leftrightarrow 12x - 20 = 14x - 21\\
 \Leftrightarrow 2x = 1\\
 \Leftrightarrow x = \frac{1}{2}\\
h,\\
\frac{4}{{x - 2}} + \frac{x}{{x + 1}} = \frac{{{x^2} - 2}}{{{x^2} - x - 2}}\,\,\,\,\,\,\left( {x \ne 2;\,\,\,x \ne  - 1} \right)\\
 \Leftrightarrow \frac{{4\left( {x + 1} \right) + x.\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x + 1} \right)}} = \frac{{{x^2} - 2}}{{\left( {x - 2} \right)\left( {x + 1} \right)}}\\
 \Leftrightarrow \frac{{{x^2} + 2x + 4}}{{\left( {x - 2} \right)\left( {x + 1} \right)}} = \frac{{{x^2} - 2}}{{\left( {x - 2} \right)\left( {x + 1} \right)}}\\
 \Leftrightarrow {x^2} + 2x + 4 = {x^2} - 2\\
 \Leftrightarrow 2x =  - 6\\
 \Leftrightarrow x =  - 3
\end{array}\)