$a/\dfrac{1}{5.8}+\dfrac{1}{8.11}+....+\dfrac{1}{x(x+5)}=\dfrac{101}{1540}$
$⇔\dfrac{1}{5}.5.(\dfrac{1}{5.8}+\dfrac{1}{8.11}+....+\dfrac{1}{x(x+5)})=\dfrac{101}{1540}$
$⇔\dfrac{1}{5}.(\dfrac{5}{5.8}+\dfrac{5}{8.11}+....+\dfrac{5}{x(x+5)})=\dfrac{101}{1540}$
$⇔\dfrac{1}{5}.(\dfrac{1}{5}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-......+\dfrac{1}{x}-\dfrac{1}{x+5}=\dfrac{101}{1540}$
$⇔\dfrac{1}{5}.(\dfrac{1}{5}-\dfrac{1}{x+5})=\dfrac{101}{1540}$
$⇒\dfrac{1}{5}-\dfrac{1}{x+5}=\dfrac{505}{1540}$
$⇔\dfrac{1}{x+5}=\dfrac{1}{5}-\dfrac{505}{1540}=\dfrac{-197}{1540}$
$⇒x+5=\dfrac{-1540}{197}$
$⇒x=\dfrac{-2525}{197} (???)$
b/
$x+\dfrac{1}{3}+ \dfrac{1}{6} + \dfrac{1}{10}+....+ \dfrac{2}{x.(x+1)}=1$
$⇔\dfrac{1}{3}+ \dfrac{1}{6} + \dfrac{1}{10}+....+ \dfrac{2}{x.(x+1)}= 1 - x$
$⇔\dfrac{2}{6} + \dfrac{2}{12} + \dfrac{2}{20} + ... + \dfrac{2}{x(x + 1)} = 1 - x$
$⇔2(\dfrac{1}{2.3} + \dfrac{1}{3.4} + \dfrac{1}{4.5} + ... + \dfrac{1}{x(x + 1)} = 1 - x$
$⇔\dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} +\dfrac{ 1}{4} - \dfrac{1}{5} + ... + \dfrac{1}{x} - \dfrac{1}{x + 1} = \dfrac{1 - x}{2}$
$⇔\dfrac{1}{2} - \dfrac{1}{x + 1)} = \dfrac{(1 - x}{2}$
$⇔\dfrac{(x + 1 - 2}{2.(x + 1} = \dfrac{(1 - x}{2}$
$⇒ (\dfrac{x - 1}{x + 1} = 1-x$
$⇔ \dfrac{1}{x + 1} = -1$
$⇔ x + 1 = -1 $
$⇒ x = -2$
