Giải hệ: \(\left\{\begin{matrix} x+y+\sqrt{x+y+3}=(x+y)^{2}+2\sqrt{x+y}\; \; \; (1)\\\sqrt{x^{2}+x+y+2} +\sqrt{x-y}=3\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; (2) \end{matrix}\right.(x,y\in R).\)
Điều kiện: \(\left\{\begin{matrix} x+y\geq 0\\ x-y\geq 0 \end{matrix}\right.(*)\)
Đặt \(t=x+y\geq 0,\) từ (1) ta có: \(t+\sqrt{t+3}=t^{2}+2\sqrt{t}\)
\(\Leftrightarrow t-t^{2}+\sqrt{t+3}-2\sqrt{t}=0\)
\(\Leftrightarrow t(1-t)+\frac{3(1-t)}{\sqrt{t+3}+2\sqrt{t}}=0\Leftrightarrow (1-t)\left ( t+\frac{3}{\sqrt{t+3}+2\sqrt{t}} \right )=0\)
\(\Leftrightarrow t=1\) (Vì \(t+\frac{3}{\sqrt{t+3}+2\sqrt{t}}>0,\forall t\geq 0\)).
Suy ra \(x+y=1\Leftrightarrow y=1-x\; \; (3)\).
Thay (3) vào (2) ta có: \(\sqrt{x^{2}+3}+\sqrt{2x-1}=3\)
\(\Leftrightarrow (\sqrt{x^{2}+3}-2)+(\sqrt{2x-1}-1)=0\Leftrightarrow \frac{x^{2}-1}{\sqrt{x^{2}+3}+2}+\frac{2x-2}{\sqrt{2x-1}+1}=0\)
\(\Leftrightarrow (x-1)\left ( \frac{x+1}{\sqrt{x^{2}+3}+2}+\frac{2}{\sqrt{2x-1}+1} \right )=0\)
\(\Leftrightarrow x=1\) (Vì \(\frac{x+1}{\sqrt{x^{2}+3}+2}+\frac{2}{\sqrt{2x-1}+1}>0,x\geq \frac{1}{2}\)).
Suy ra (x = 1; y = 0), thỏa mãn (*).
Vậy hệ đã cho có nghiệm duy nhất (x = 1; y = 0).
