Lời giải:ĐKXĐ: $x\neq 3; y\geq -1$Đặt $\frac{1}{x-3}=a; \sqrt{y+1}=b(b\geq 0)$ thì hpt trở thành:$\left\{\begin{matrix} a+3b=5\\ 2a-5b=-1\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 2a+6b=10\\ 2a-5b=-1\end{matrix}\right.$$\Rightarrow (2a+6b)-(2a-5b)=11$$\Leftrightarrow 11b=11$$\Leftrightarrow b=1$ (tm) $a=5-3b=5-3=2$ Khi đó: $(a,b)=(2,1)$$\Leftrightarrow (\frac{1}{x-3}, \sqrt{y+1})=(2,1)$$\Leftrightarrow (x,y)=(\frac{7}{2}, 0)$Câu trả lời: Lời giải:ĐKXĐ: $x\neq 3; y\geq -1$Đặt $\frac{1}{x-3}=a; \sqrt{y+1}=b(b\geq 0)$ thì hpt trở thành:$\left\{\begin{matrix} a+3b=5\\ 2a-5b=-1\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 2a+6b=10\\ 2a-5b=-1\end{matrix}\right.$$\Rightarrow (2a+6b)-(2a-5b)=11$$\Leftrightarrow 11b=11$$\Leftrightarrow b=1$ (tm) $a=5-3b=5-3=2$ Khi đó: $(a,b)=(2,1)$$\Leftrightarrow (\frac{1}{x-3}, \sqrt{y+1})=(2,1)$$\Leftrightarrow (x,y)=(\frac{7}{2}, 0)$