Giai truong hop tong quat luon:Dieu kien: $x \ge 2;y \ge 1.$ Tu PT $(2)$ suy ra $m \ge 0.$ He da cho tuong duong voi:
$\begin{cases}\sqrt{x+1}+\sqrt{x-2}+\sqrt{y+2}+\sqrt{y-1}=2m+2 \\ \sqrt{x+1}-\sqrt{x-2}+\sqrt{y+2}-\sqrt{y-1}=2 \end{cases}$
Dat $\begin{cases}a=\sqrt{x+1}+\sqrt{x-2} (a>0) \\ b=\sqrt{y+2}+\sqrt{y-1} (b>0) \end{cases},$ ta co:
$\begin{cases}\sqrt{x+1}-\sqrt{x-2}=\frac{3}{a} \\ \sqrt{y+2}-\sqrt{y-1}=\frac{3}{b} \end{cases},$
Khi do he da cho tro thanh:
$\begin{cases}a+b=2m+2 \\ \frac{3}{a}+\frac{3}{b}=2 \end{cases}$
$\Leftrightarrow \begin{cases}a+b=2m+2 \\ ab=3(m+1) \end{cases}$
Theo dinh li Viete dao, he da cho co nghiem khi va chi khi:
$S^2 \ge 4P\Leftrightarrow (2m+2)^2 \ge 4.3(m+1)\Leftrightarrow m \ge 2.$
KL: he PT da cho co nghiem khi va chi khi $m \ge 2.$
Ap dung voi $m=2,$ ta co:
$\begin{cases}a+b=6 \\ ab=9 \end{cases}\Leftrightarrow a=b=3$ (vi $a,b>0$)
$\Leftrightarrow \begin{cases} \sqrt{x+1}+\sqrt{x-2}=3\\ \sqrt{y+2}+\sqrt{y-1}=3 \end{cases}$ (giai 2 PT nay bang PP lien hop hay binh phuong tuy em.)
$\Leftrightarrow \begin{cases}x=3 \\ y=2 \end{cases}.$
Neu thay hay thi vote manh vao...:))