$\dfrac{1}{x(x+3)}+\dfrac{1}{(x+3)(x+6)}+\dfrac{1}{(x+6)(x+9)}+\dfrac{1}{(x+9)(x+12)}$
$=\dfrac{1}{3}\Bigg(\dfrac{1}{x}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+9}+\dfrac{1}{x+9}-\dfrac{1}{x+12}\Bigg)$
$=\dfrac{1}{3}\Bigg(\dfrac{1}{x}-\dfrac{1}{x+12}\Bigg)$
$=\dfrac{1}{3}.\dfrac{12}{x(x+12)}$
$=\dfrac{4}{x(x+12)}$