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\begin{cases}\sqrt{x+1} + \sqrt{y-1} = 4 \\ \sqrt{x+6} + \sqrt{y+4} = 6 \end{cases} $\Leftrightarrow$ \begin{cases}x + y + 2\sqrt{(x+1)(y-1)} = 16 \\ x + y + 10 + 2\sqrt{(x+6)(y+4)}= 36 \end{cases} $\Leftrightarrow $ \begin{cases} 2\sqrt{(x+1)(y-1)} = 16-x-y (1) \\ 2\sqrt{(x+6)(y+4)} - 10 = 16 - x - y (2)\end{cases} Lấy (1) trừ (2) theo vế, ta được :
$\sqrt{(x+6)(y+4)}$ = $\sqrt{(x+1)(y-1)}$ + 5
$\Leftrightarrow$ $(x+6)(y+4) = (x+1)(y-1) + 25 + 10\sqrt{(x+1)(y-1)}$
$\Leftrightarrow$ $ xy + 4x + 6y + 24 = xy - x + y + 24 + 10\sqrt{(x+1)(y-1)}$
$\Leftrightarrow$ $ x + y = 2\sqrt{(x+1)(y-1)}$
$\Leftrightarrow$ $ x^2 + y^2 + 2xy = 4xy - 4x + 4y - 4$
$\Leftrightarrow$ $ x^2 + y^2 - 2xy + 4x - 4y + 4 = 0$
$\Leftrightarrow$ $ x^2 - 2x(y-2) + (y-2)^2 = 0$
$\Leftrightarrow$ $ (x-y+2)^2 = 0$
$\Leftrightarrow$ $ x = y - 2 (3) $
Thế $(3) $ vào $(1) $ ta có
$(1) \Leftrightarrow \sqrt{y-1} = 2 \Leftrightarrow y = 5 \Rightarrow x=3 $
Vậy $ (x;y) = (3;5) $
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