Đáp án đúng: A
Giải chi tiết:\(\begin{align} & M={{x}^{2}}+{{y}^{2}}+\frac{3}{x+y+1}={{x}^{2}}+{{y}^{2}}+1+2xy+2x+2y-2(x+y+1)+\frac{3}{x+y+1} \\ & \,\,\,\,\,\,\,={{\left( x+y+1 \right)}^{2}}-2\left( x+y+1 \right)+\frac{3}{x+y+1} \\ \end{align}\)
Đặt \(t=x+y+1\ge 2\sqrt{xy}+1=3\)
\(\begin{align} & \Rightarrow M={{t}^{2}}-2t+\frac{3}{t} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,={{\left( t-3 \right)}^{2}}+4t+\frac{3}{t}-9 \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,={{\left( t-3 \right)}^{2}}+\frac{t}{3}+\frac{3}{t}+\frac{11}{3}t-9 \\ \end{align}\)
Áp dụng bất đẳng thức Cô – si ta có \(\frac{t}{3}+\frac{3}{t}\ge 2\sqrt{\frac{t}{3}.\frac{3}{t}}=2\) \(\Rightarrow M\ge 0+2+\frac{11}{3}.3-9=4\)
Dấu “=” xảy ra khi \(\left\{ \begin{align} & x=y \\ & t=3 \\ & \frac{t}{3}=\frac{3}{t} \\ \end{align} \right.\Leftrightarrow x=y=1\)
Vậy \({{M}_{\min }}=4\Leftrightarrow x=y=1\)
Chọn A