Tìm x: $\frac{3}{1X5}$ + $\frac{3}{5X9}$ +…+ $\frac{3}{(4x+1)X(4x+5)}$ Lưu ý: Dâu x lớn ( X ) là dấu nhân Dấu x nhỏ ( x ) là chữ số x

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)}$

$= \dfrac{3}{4}.(\dfrac{1}{1}-\dfrac{1}{5})+ \dfrac{3}{4}.(\dfrac{1}{5}-\dfrac{1}{9})+…+ \dfrac{3}{4}.(\dfrac{1}{4x+1}-\dfrac{1}{4x+5})$

$= \dfrac{3}{4}.(\dfrac{1}{1}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+…+\dfrac{1}{4x+1}-\dfrac{1}{4x+5})$

$= \dfrac{3}{4}.(1-\dfrac{1}{4x+5})$