Đáp án:$(x;y)=(0;1)$
Giải thích các bước giải:
\(\begin{array}{l}
\left\{ \begin{array}{l}
(4y - 1)\sqrt {{x^2} + 1} - 2y = 2{x^2} + 1(1)\\
{x^2}y + {y^2} = 1(2)
\end{array} \right.\\
(1) \Leftrightarrow 2(2y - 1)\sqrt {{x^2} + 1} + \sqrt {{x^2} + 1} - 2y + 1 = 2({x^2} + 1)\\
\Leftrightarrow \left[ {2(2y - 1)\sqrt {{x^2} + 1} - (2y - 1)} \right] + \left[ {\sqrt {{x^2} + 1} - 2({x^2} + 1)} \right] = 0\\
\Leftrightarrow (2y - 1)\left( {2\sqrt {{x^2} + 1} - 1} \right) - \sqrt {{x^2} + 1} (2\sqrt {{x^2} + 1} - 1) = 0\\
\Leftrightarrow (2y - 1 - \sqrt {{x^2} + 1} )(2\sqrt {{x^2} + 1} - 1) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
2y - 1 = \sqrt {{x^2} + 1} \Rightarrow \left\{ \begin{array}{l}
4{y^2} - 4y + 1 = {x^2} + 1\\
y \ge \frac{1}{2}
\end{array} \right.\\
2\sqrt {{x^2} + 1} = 1 \to vn
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
y \ge \frac{1}{2}\\
{x^2} = 4{y^2} - 4y
\end{array} \right.(*)\\
Thay\;(*)\;vao\;(2)\\
\Rightarrow \left\{ \begin{array}{l}
y(4{y^2} - 4y) + {y^2} = 1\\
y \ge \frac{1}{2}
\end{array} \right.\\
\Rightarrow y = 1 \Rightarrow x = 0\\
\end{array}\)
