\(\Leftrightarrow\dfrac{3}{x\left(x+3\right)}+\dfrac{3}{\left(x+3\right)\left(x+6\right)}+...+\dfrac{3}{\left(x+9\right)\left(x+12\right)}=\dfrac{3}{16}\)
=>\(\dfrac{1}{x}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+6}+...+\dfrac{1}{x+9}-\dfrac{1}{x+12}=\dfrac{3}{16}\)=>\(\dfrac{1}{x}-\dfrac{1}{x+12}=\dfrac{3}{16}\)
=>\(\dfrac{x+12-x}{x\left(x+12\right)}=\dfrac{3}{16}\)
=>12/x(x+12)=3/16
=>4/x(x+12)=1/16
=>x(x+12)=64
=>x^2+12x-64=0
=>x^2+16x-4x-64=0
=>(x+16)(x-4)=0
=>x=4 hoặc x=-16