$`<=>\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{3}{10}$
$<=>\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{3}{10}.$
$<=>\dfrac{1}{3}+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+\left(\dfrac{1}{6}-\dfrac{1}{6}\right)+...+\left(\dfrac{1}{x}-\dfrac{1}{x}\right)-\dfrac{1}{x+1}=\dfrac{3}{10}$.
$<=>\dfrac{1}{3}+0+0+0+...+0-\dfrac{1}{x+1}=\dfrac{3}{10}.$
$<=>\dfrac{1}{3}-\dfrac{1}{x+1}=\dfrac{3}{10}.$
$<=>\dfrac{1}{x+1}=\dfrac{1}{3}-\dfrac{3}{10}.$
$<=>\dfrac{1}{x+1}=\dfrac{10}{30}-\dfrac{9}{30}.$
$<=>\dfrac{1}{x+1}=\dfrac{1}{30}.$
$\Rightarrow x+1=30.$
$\Rightarrow x=30-1=29.$
$Vậy x=29.$
