Ans: A
$\begin{array}{rcl}
\dfrac{5}{4k+3} – \dfrac{2}{4k-3} & = & \dfrac{5(4k-3)}{(4k+3)(4k-3)} – \dfrac{2(4k+3)}{(4k-3)(4k+3)} \\
& = & \dfrac{5(4k-3) – 2(4k+3)}{(4k+3)(4k-3)} \\
& = & \dfrac{20k – 15 – 8k -6}{16k^2 – 9} \\
& = & \dfrac{12k – 21}{16k^2 -9}
\end{array}$
$\begin{array}{rcl}
\dfrac{5}{4k+3} – \dfrac{2}{4k-3} & = & \dfrac{5(4k-3)}{(4k+3)(4k-3)} – \dfrac{2(4k+3)}{(4k-3)(4k+3)} \\
& = & \dfrac{5(4k-3) – 2(4k+3)}{(4k+3)(4k-3)} \\
& = & \dfrac{20k – 15 – 8k -6}{16k^2 – 9} \\
& = & \dfrac{12k – 21}{16k^2 -9}
\end{array}$
