ĐKXĐ: \(\sqrt{x^2+2x+2}-\left(x+1\right)\ge0\)
\(\Rightarrow\sqrt{x^2+2x+2}\ge x+1\)
Ta có \(\sqrt{x^2+2x+2}=\sqrt{\left(x+1\right)^2+1}>\sqrt{\left(x+1\right)^2}=\left|x+1\right|\ge x+1\)
\(\Rightarrow\sqrt{x^2+2x+2}-\left(x+1\right)>0\) \(\forall x\)
\(\Rightarrow D=R\)