Giải thích các bước giải:
Ta có:
$\begin{split}\dfrac{3}{(4x+1)(4x+5)}&=\dfrac{3}{4}.\dfrac{1}{(4x+1)(4x+5)}\\&=\dfrac{3}{4}.\dfrac{4x+5-(4x+1)}{(4x+1)(4x+5)}\\&=\dfrac{3}{4}(\dfrac{4x+5}{(4x+1)(4x+5)}-\dfrac{4x+1}{(4x+1)(4x+5)})\\&=\dfrac{3}{4}(\dfrac{1}{4x+1}-\dfrac{1}{4x+5})\end{split}$
$ \dfrac{3}{1.5}+\dfrac{3}{5.9}+...+\dfrac{3}{(4x+1)(4x+5)}$
$= \dfrac{3}{4}.(\dfrac{1}{1}-\dfrac{1}{5})+ \dfrac{3}{4}.(\dfrac{1}{5}-\dfrac{1}{9})+...+ \dfrac{3}{4}.(\dfrac{1}{4x+1}-\dfrac{1}{4x+5})$
$= \dfrac{3}{4}.(\dfrac{1}{1}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{4x+1}-\dfrac{1}{4x+5})$
$= \dfrac{3}{4}.(1-\dfrac{1}{4x+5})$