Hpt : $\begin{cases}3(2-x)\sqrt{2-y^{2}}=2-y+\frac{4}{x+1} (1)\\ (x^{2}+xy-x+y-2)\sqrt{2-y^{2}}+2=x+y (2)\end{cases}$ĐK : $x\neq 0, y\in [-\sqrt{2};\sqrt{2}]$$(2)\Leftrightarrow (x+1)(x+y-2)\sqrt{2-y^{2}}=x+y-2$ $(*)$TH1: $x+y-2=0\rightarrow x=2-y$ Thế vào pt (1) : $3y\sqrt{2-y^{2}}=2-y+\frac{4}{3-y}$$\Leftrightarrow 3y\sqrt{2-y^{2}}-3y=2-2y+\frac{4}{3-y}-2y$$\Leftrightarrow 3y\frac{(1-y)(1+y)}{\sqrt{2-y^{2}}+1}-2(1-y)- \frac{2(2-y)(1-y)}{3-y}=0$$ \Leftrightarrow (1-y)\left[ {...}] \right.=0$$\Leftrightarrow x=y=1$ (tm)TH2 :$x+y-2 \neq 0$ $(*) \Leftrightarrow (x+1)\sqrt{2-y^{2}}=1 \rightarrow ...... $Vậy hpt có nghiệm : $(x;y)=(1;1)$

Hpt : $\begin{cases}3(2-x)\sqrt{2-y^{2}}=2-y+\frac{4}{x+1} (1)\\ (x^{2}+xy-x+y-2)\sqrt{2-y^{2}}+2=x+y (2)\end{cases}$ĐK : $x\neq 0, y\in [-\sqrt{2};\sqrt{2}]$$(2)\Leftrightarrow (x+1)(x+y-2)\sqrt{2-y^{2}}=x+y-2$ $(*)$TH1: $x+y-2=0\rightarrow x=2-y$ Thế vào pt (1) : $3y\sqrt{2-y^{2}}=2-y+\frac{4}{3-y}$$\Leftrightarrow 3y\sqrt{2-y^{2}}-3y=2-2y+\frac{4}{3-y}-2y$$\Leftrightarrow 3y\frac{(1-y)(1+y)}{\sqrt{2-y^{2}}+1}-2(1-y)- \frac{2(2-y)(1-y)}{3-y}=0$$ \Leftrightarrow (1-y)\left[ {...}] \right.=0$$\Leftrightarrow x=y=1$ (tm)TH2 :$x+y-2 \neq 0$ $(*) \Leftrightarrow (x+1)\sqrt{2-y^{2}}=1 \rightarrow ...... $Vậy hpt có nghiệm : $(x;y)=(1;1)$ .... ( có thể còn nhưng bạn tự tìm đi :v )