Đk: $\begin{cases}-2\leq x\leq 4\\ y\geq 1 \\ x+3y+2\geq 0\end{cases}$Pt $(1): 2\sqrt{x+3y+2}=\sqrt{x+2}+3\sqrt{y}$Bình phương 2 vế ta được: $4(x+3y+2)=x+3y+2+6y+6\sqrt{y(x+2)}$$\Leftrightarrow x+2-2\sqrt{y(x+2})+y=0$$\Leftrightarrow (\sqrt{x+2}-\sqrt{y})^2=0$$\Leftrightarrow\sqrt{x+2}=\sqrt{y}$$\Leftrightarrow x+2=y$Thế vào $(2)\Leftrightarrow \sqrt{x+1}-\sqrt{4-x}+8-x^2=0$$\Leftrightarrow \sqrt{x+1}-2+1-\sqrt{4-x}+9-x^2=0$$\Leftrightarrow \frac{x-3}{\sqrt{x+1}+2}+\frac{x-3}{\sqrt{4-x}+1}-(x-3)(x+3)=0$$\Leftrightarrow x=3\vee \frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{4-x}+1}-x-3=0$ $(\bigstar)$$(\bigstar)<0\Rightarrow (\bigstar)$ vô nghiệmTừ đó $ x=3\Rightarrow y=5$
Đk: $\begin{cases}-2\leq x\leq 4\\ y\geq 1 \\ x+3y+2\geq 0\end{cases}$Pt $(1): 2\sqrt{x+3y+2}=\sqrt{x+2}+3\sqrt{y}$Bình phương 2 vế ta được: $4(x+3y+2)=x+3y+2+6y+6\sqrt{y(x+2)}$$\Leftrightarrow x+2-2\sqrt{y(x+2})+y=0$$\Leftrightarrow (\sqrt{x+2}-\sqrt{y})^2=0$$\Leftrightarrow\sqrt{x+2}=\sqrt{y}$$\Leftrightarrow x+2=y$Thế vào $(2)\Leftrightarrow \sqrt{x+1}-\sqrt{4-x}+8-x^2=0$Pt này vô nghiệm $\Leftrightarrow $ Hệ vô nghiệm
Đk: $\begin{cases}-2\leq x\leq 4\\ y\geq 1 \\ x+3y+2\geq 0\end{cases}$Pt $(1): 2\sqrt{x+3y+2}=\sqrt{x+2}+3\sqrt{y}$Bình phương 2 vế ta được: $4(x+3y+2)=x+3y+2+6y+6\sqrt{y(x+2)}$$\Leftrightarrow x+2-2\sqrt{y(x+2})+y=0$$\Leftrightarrow (\sqrt{x+2}-\sqrt{y})^2=0$$\Leftrightarrow\sqrt{x+2}=\sqrt{y}$$\Leftrightarrow x+2=y$Thế vào $(2)\Leftrightarrow \sqrt{x+1}-\sqrt{4-x}+8-x^2=0$
$\Left
rightarrow \sqrt{x+1}-2+1-\sqrt{4-x}+9-x^2=0$$\Leftrigh
tarrow \frac{x-3}{\sqrt{x+1}+2}+\frac{x-3}{\sqrt{4-x}+1}-(x-3)(x+3)=0$$\Leftrightarrow
x=3\vee \frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{4-x}+1}-x-3=0$
$(\bigstar)$$(\bigstar)<0\Rightarrow (\bigstar)$ vô nghiệm
Từ đó $ x=3\Rightarrow y=5$