Đáp án:
$S =\{(-1;1)\}$
Giải thích các bước giải:
$\quad \begin{cases}x - 2y^2 = -3\\\sqrt{x+2} + 2y = 3\end{cases}$
$\Leftrightarrow \begin{cases}x\geqslant - 2\\x = 2y^2 - 3\\\sqrt{2y^2 - 1} = 3-2y\end{cases}$
$\Leftrightarrow \begin{cases}x\geqslant -2\\y^2 \geqslant \dfrac12\\x = 2y^2 -3\\3 - 2y \geqslant 0\\2y^2 - 1 = (3-2y)^2\end{cases}$
$\Leftrightarrow \begin{cases}x\geqslant -2\\\left[\begin{array}{l}\dfrac{\sqrt2}{2}\leqslant y \leqslant \dfrac32\\y\leqslant -\dfrac{\sqrt2}{2}\end{array}\right.\\x = 2y^2 - 3\\y^2 - 6y + 5 = 0\end{cases}$
$\Leftrightarrow \begin{cases}x\geqslant -2\\\left[\begin{array}{l}\dfrac{\sqrt2}{2}\leqslant y \leqslant \dfrac32\\y\leqslant -\dfrac{\sqrt2}{2}\end{array}\right.\\x = 2y^2 - 3\\\left[\begin{array}{l}y = 1\\y = 5\end{array}\right.\end{cases}$
$\Leftrightarrow \begin{cases}x\geqslant -2\\x = 2y^2 -3\\y = 1\end{cases}$
$\Leftrightarrow \begin{cases}x = -1\\y = 1\end{cases}$
Vậy $S =\{(-1;1)\}$
