Giải thích các bước giải:
c,
ĐKXĐ: \(x + y > 0\)
Ta có:
\(\begin{array}{l}
\left\{ \begin{array}{l}
{x^2} + {y^2} + \frac{{2xy}}{{x + y}} = 1\,\,\,\,\,\,\,\left( 1 \right)\\
\sqrt {x + y} = {x^2} - y\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow \left( {{x^2} + {y^2} + 2xy} \right) + \frac{{2xy}}{{x + y}} - 2xy = 1\\
\Leftrightarrow {\left( {x + y} \right)^2} - 1 + 2xy\left( {\frac{1}{{x + y}} - 1} \right) = 0\\
\Leftrightarrow \left( {x + y - 1} \right)\left( {x + y + 1} \right) + 2xy\left( {\frac{{x + y - 1}}{{x + y + 1}}} \right) = 0\\
\Leftrightarrow \left( {x + y - 1} \right)\left( {x + y + 1 + \frac{{2xy}}{{x + y + 1}}} \right) = 0\\
x + y > 0 \Rightarrow x + y - 1 = 0\\
\Leftrightarrow y = 1 - x\\
\left( 2 \right) \Leftrightarrow \sqrt {x + 1 - x} = {x^2} - \left( {1 - x} \right)\\
\Leftrightarrow 1 = {x^2} - 1 + x\\
\Leftrightarrow {x^2} + x - 2 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1 \Rightarrow y = 0\,\,\,\,\left( {t/m} \right)\\
x = - 2 \Rightarrow y = 3\,\,\,\left( {t/m} \right)
\end{array} \right.
\end{array}\)
