a: \(\dfrac{1}{9\cdot13}+\dfrac{1}{13\cdot17}+...+\dfrac{1}{\left(4x+1\right)\left(4x+5\right)}=\dfrac{10}{441}\)
=>\(\dfrac{4}{9\cdot13}+\dfrac{4}{13\cdot17}+...+\dfrac{4}{\left(4x+1\right)\left(4x+5\right)}=\dfrac{40}{441}\)
=>\(\dfrac{1}{9}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{17}+...+\dfrac{1}{4x+1}-\dfrac{1}{4x+5}=\dfrac{40}{441}\)
=>\(\dfrac{1}{9}-\dfrac{1}{4x+5}=\dfrac{40}{441}\)
=>\(\dfrac{1}{4x+5}=\dfrac{1}{9}-\dfrac{40}{441}=\dfrac{1}{49}\)
=>4x+5=49
=>4x=44
=>x=11
b: \(\dfrac{1}{3\cdot10}+\dfrac{1}{10\cdot17}+...+\dfrac{1}{\left(7x-4\right)\left(7x+3\right)}\)
\(=\dfrac{1}{7}\left(\dfrac{7}{3\cdot10}+\dfrac{7}{10\cdot17}+...+\dfrac{7}{\left(7x-4\right)\left(7x+3\right)}\right)\)
\(=\dfrac{1}{7}\left(\dfrac{1}{3}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{17}+...+\dfrac{1}{7x-4}-\dfrac{1}{7x+3}\right)\)
\(=\dfrac{1}{7}\left(\dfrac{1}{3}-\dfrac{1}{7x+3}\right)\)
\(=\dfrac{1}{7}\cdot\dfrac{7x+3-3}{3\left(7x+3\right)}=\dfrac{x}{3\left(7x+3\right)}\)
