Nghiệm của hệ phương trình \(\left\{\begin{array}{l} \sqrt{x+3}-2 \sqrt{y+1}=2 \\ 2 \sqrt{x+3}+\sqrt{y+1}=4 \end{array}\right.\) .

* Hướng dẫn giải

 \(\begin{aligned} &\text { Điều kiện: } x \geq-3 ; y \geq-1 \text { Ta có }\left\{\begin{array}{l} \sqrt{x+3}-2 \sqrt{y+1}=2 \\ 2 \sqrt{x+3}+\sqrt{y+1}=4 \end{array}\right.\\ &\Leftrightarrow\left\{\begin{array}{l} 2 \sqrt{x+3}-4 \sqrt{y+1}=4 \\ 2 \sqrt{x+3}+\sqrt{y+1}=4 \end{array} \Leftrightarrow\left\{\begin{array}{l} \sqrt{x+3}-2 \sqrt{y+1}=2 \\ -5 \sqrt{y+1}=0 \end{array}\right.\right.\\ &\Leftrightarrow\left\{\begin{array}{l} y=-1 \\ \sqrt{x+3}-2 \cdot \sqrt{(-1)+1}=2 \end{array} \Leftrightarrow\left\{\begin{array}{l} y=-1 \\ \sqrt{x+3}=2 \end{array} \Leftrightarrow\left\{\begin{array}{l} y=-1 \\ x+3=4 \end{array} \Leftrightarrow\left\{\begin{array}{l} y=-1 \\ x=1 \end{array}(t m)\right.\right.\right.\right. \end{aligned}\)