Đáp án:
\(S = \left\{(-2;1);\left(1;-\dfrac12\right);(1;1);(\sqrt2 - 1;1 - \sqrt2);(-1-\sqrt2;1 + \sqrt2) \right\}\)
Giải thích các bước giải:
\(\begin{array}{l}
\quad \begin{cases}x\left(x + \dfrac1y\right) = 2y\qquad (*)\\
x^2 + \dfrac{1}{y^2} + x + \dfrac{1}{y} = 4\end{cases}\quad (y \ne 0)\\
\Leftrightarrow \begin{cases}\dfrac{2x}{y}\left(x + \dfrac1y\right) = 4\\
\left(x + \dfrac{1}{y}\right)^2 + x + \dfrac{1}{y} - \dfrac{2x}{y} = 4\end{cases}\\
\text{Trừa vế theo vế ta được:}\\
\left(x + \dfrac1y\right)^2 + \left(x + \dfrac1y\right) - \dfrac{2x}{y} - \dfrac{2x}{y}\left(x + \dfrac1y\right) = 0\\
\Leftrightarrow \left(x + \dfrac1y\right)\left(x + \dfrac1y + 1\right) - \dfrac{2x}{y}\left(x + \dfrac1y+1\right) = 0\\
\Leftrightarrow \left(x + \dfrac1y + 1\right)\left(x + \dfrac1y - \dfrac{2x}{y}\right)=0\\
\Leftrightarrow \left[\begin{array}{l}x + \dfrac1y = -1\\x + \dfrac{1}{y} = \dfrac{2x}{y}\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = - 1 - \dfrac{1}{y}\\x = \dfrac{1}{2 - y}\end{array}\right.\\
+)\quad \text{Với $x = - 1 - \dfrac{1}{y}$, thay vào $(*)$ ta được:}\\
\left(- 1 - \dfrac1y\right)\cdot (-1) = 2y\\
\Leftrightarrow 2y^2 - y - 1 =0\\
\Leftrightarrow \left[\begin{array}{l}y = 1\quad\Rightarrow x = -2\\y = -\dfrac12\Rightarrow x = 1\end{array}\right.\\
+)\quad \text{Với $x = \dfrac{1}{2-y}$, thay vào $(*)$ ta được:}\\
\quad \dfrac{1}{2-y}\left(\dfrac{1}{2-y} + \dfrac{1}{y}\right) = 2y\\
\Leftrightarrow y^4 - 4y^3 + 4y^2 - 1 =0\\
\Leftrightarrow (y-1)^2(y^2 - 2y - 1) = 0\\
\Leftrightarrow \left[\begin{array}{l}y = 1\qquad\ \ \Rightarrow x = 1\\y = 1 - \sqrt2\Rightarrow x = \sqrt2 - 1\\y = 1 + \sqrt2\Rightarrow -1 - \sqrt2\end{array}\right.\\
\text{Vậy}\ S = \left\{(-2;1);\left(1;-\dfrac12\right);(1;1);(\sqrt2 - 1;1 - \sqrt2);(-1-\sqrt2;1 + \sqrt2) \right\}
\end{array}\)
