\begin{cases}\sqrt{x+1}+y=\sqrt{y+1}+x \\ x^{2}+2xy-y^{2}=4 \end{cases}
Đáp án đây nha, mong là đúng hihi!!!
Điều kiện: $x, y\geq -1$
Ta có: $pt(1)\Leftrightarrow (\sqrt{x+1}-\sqrt{y+1})(\sqrt{x+1}+\sqrt{y+1}-1)=0$
TH1: $\sqrt{x+1}=\sqrt{y+1}$ thì x=y, thay vào pt(2) là OK
TH2: $\sqrt{x+1}+\sqrt{y+1}=1$  (*)
Mặt khác, ta thấy: $pt(2)\Leftrightarrow (x+y)^2=4+2y^2\geq 4$
$\Rightarrow x+y\geq 2$ (**) (vì $x, y\geq -1$)
Từ (*) ta có: $\sqrt{x+1}\leq 1\Leftrightarrow x\leq 0$, tương tự: $y\leq 0$. Do đó $x+y\leq 0$ ( trái với (**) nên TH này loại)
Vậy...
Có gì thắc mắc bảo mình nha, đúng tích giùm mình, hihi!!!

Bạn cần đăng nhập để có thể gửi đáp án

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