\(\frac{1}{x\left(x-1\right)}+\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}=\frac{3}{4}.\)
\(\Rightarrow\frac{1}{x-1}-\frac{1}{x}+\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}=\frac{3}{4}\)
\(\Rightarrow\frac{1}{x-1}-\frac{1}{x+2}=\frac{3}{4}\)
\(\Leftrightarrow4\left(x+2\right)-4\left(x-1\right)=3\left(x+2\right)\left(x-1\right)\)
\(\Leftrightarrow4x+8-4x+4=3x^2+3x-6\)
\(\Leftrightarrow3x^2+3x-18=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Vậy nghiệm nhỏ nhất là \(x=-3.\)