\(\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}=\dfrac{3}{4}\)

\(\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}=\dfrac{3}{4}\)

\(\dfrac{1}{x+1}-\dfrac{1}{x+4}=\dfrac{3}{4}\)

\(\dfrac{\left(x+4\right)-\left(x+1\right)}{\left(x+1\right)\left(x+4\right)}=\dfrac{3}{4}\)

\(\dfrac{3}{\left(x+1\right)\left(x+4\right)}=\dfrac{3}{4}\)

=> (x + 1)(x + 4) = 4

=> x² + 5x + 4 = 4

=> x² + 5x = 0

=> x(x + 5) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

Hence , x = 0 or x = -5