Ai đó giúp em bài 5 với ạ???? em đang gấp lắm

Giải thích các bước giải:

Ta có:

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