ĐKXĐ:\(x\ne-1,x\ne-2\)
\(\dfrac{\left(x+2\right)\left(2x-1\right)-x-2}{x^2+3x+2}=0\\ \Rightarrow2x^2+4x-x-2-x-2=0\\ \Leftrightarrow2x^2+2x-4=0\\ \Leftrightarrow x^2+x-2=0\\ \Leftrightarrow\left(x+2\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\left(ktm\right)\\x=1\left(tm\right)\end{matrix}\right.\)
Vậy \(x\in\left\{1\right\}\)
Chọn D