Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
\dfrac{{3{x^2} - 2x - 5}}{A} = \dfrac{{3x - 5}}{{2x - 3}}\\
 \Leftrightarrow \dfrac{{3{x^2} - 2x - 5}}{A} = \dfrac{{\left( {3x - 5} \right)\left( {x + 1} \right)}}{{\left( {2x - 3} \right)\left( {x + 1} \right)}}\\
 \Leftrightarrow \dfrac{{3{x^2} - 2x - 5}}{A} = \dfrac{{3{x^2} - 2x - 5}}{{2{x^2} - x - 3}}\\
 \Leftrightarrow A = 2{x^2} - x - 3\\
b,\\
\dfrac{{2{x^2} + 3x - 2}}{{{x^2} - 4}} = \dfrac{B}{{{x^2} - 4x + 4}}\\
 \Leftrightarrow \dfrac{{\left( {x + 2} \right)\left( {2x - 1} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{B}{{{x^2} - 4x + 4}}\\
 \Leftrightarrow \dfrac{{2x - 1}}{{x - 2}} = \dfrac{B}{{{x^2} - 4x + 4}}\\
 \Leftrightarrow \dfrac{{\left( {2x - 1} \right)\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x - 2} \right)}} = \dfrac{B}{{{x^2} - 4x + 4}}\\
 \Leftrightarrow \dfrac{{2{x^2} - 5x + 2}}{{{x^2} - 4x + 4}} = \dfrac{B}{{{x^2} - 4x + 4}}\\
 \Leftrightarrow B = 2{x^2} - 5x + 2\\
c,\\
\dfrac{C}{{2x + 1}} = \dfrac{{10{x^2} - 5x}}{{4{x^2} - 1}}\\
 \Leftrightarrow \dfrac{C}{{2x + 1}} = \dfrac{{5x.\left( {2x - 1} \right)}}{{\left( {2x - 1} \right)\left( {2x + 1} \right)}}\\
 \Leftrightarrow \dfrac{C}{{2x + 1}} = \dfrac{{5x}}{{2x + 1}}\\
 \Rightarrow C = 5x\\
d,\\
\dfrac{{4{x^2} - 16x + 16}}{{{x^2} - 4}} = \dfrac{D}{{x + 2}}\\
 \Leftrightarrow \dfrac{{4.\left( {{x^2} - 4x + 4} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{D}{{x + 2}}\\
 \Leftrightarrow \dfrac{{4.{{\left( {x - 2} \right)}^2}}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{D}{{x + 2}}\\
 \Leftrightarrow \dfrac{{4.\left( {x - 2} \right)}}{{x + 2}} = \dfrac{D}{{x + 2}}\\
 \Leftrightarrow D = 4\left( {x - 2} \right)
\end{array}\)