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Đáp án:
\[\left\{ \begin{array}{l}
x = \frac{{17 - \sqrt {33} }}{8}\\
y = \frac{{ - 9 + \sqrt {33} }}{8}
\end{array} \right.\]
Giải thích các bước giải:
ĐKXĐ: \(\left\{ \begin{array}{l}
x + y \ge 1\\
x > 0\\
y \ne 0\\
x \ne - y
\end{array} \right.\)
Ta có:
\(\begin{array}{l}
\left\{ \begin{array}{l}
\frac{{{{\left( {x + y} \right)}^2} - 1}}{{xy}} + \frac{2}{{x + y}} = 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\\
4{x^2} + 5y - 13 + \sqrt {x + y - 1} + 6\sqrt x = 0\,\,\,\,\,\left( 2 \right)
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow \frac{{\left( {x + y - 1} \right)\left( {x + y + 1} \right)}}{{xy}} + 2.\left( {\frac{1}{{x + y}} - 1} \right) = 0\\
\Leftrightarrow \frac{{\left( {x + y - 1} \right)\left( {x + y + 1} \right)}}{{xy}} + 2.\frac{{1 - \left( {x + y} \right)}}{{x + y}} = 0\\
\Leftrightarrow \frac{{\left( {x + y - 1} \right)\left( {x + y + 1} \right)}}{{xy}} - 2.\frac{{x + y - 1}}{{x + y}} = 0\\
\Leftrightarrow \left( {x + y - 1} \right).\left[ {\frac{{x + y + 1}}{{xy}} - \frac{2}{{x + y}}} \right] = 0\\
\Leftrightarrow \left( {x + y - 1} \right).\left[ {\frac{{{{\left( {x + y} \right)}^2} + x + y - 2xy}}{{xy\left( {x + y} \right)}}} \right] = 0\\
\Leftrightarrow \left( {x + y - 1} \right).\left( {{x^2} + {y^2} + x + y} \right) = 0\\
x + y \ge 1 \Rightarrow {x^2} + {y^2} + x + y \ge 1 > 0\\
\Rightarrow x + y - 1 = 0\\
\Leftrightarrow y = 1 - x\\
\left( 2 \right) \Leftrightarrow 4{x^2} + 5y - 13 + \sqrt {x + y - 1} + 6\sqrt x = 0\\
\Leftrightarrow 4{x^2} + 5.\left( {1 - x} \right) - 13 + \sqrt {1 - 1} + 6\sqrt x = 0\\
\Leftrightarrow 4{x^2} + 5 - 5x - 13 + 6\sqrt x = 0\\
\Leftrightarrow 4{x^2} - 5x + 6\sqrt x - 8 = 0\,\,\,\,\,\,\,\,\,\,\left( 3 \right)\\
t = \sqrt x \,\,\,\,\left( {t > 0} \right) \Rightarrow x = {t^2}\\
\left( 3 \right) \Leftrightarrow 4{t^4} - 5{t^2} + 6t - 8 = 0\\
\Leftrightarrow \left( {4{t^4} + 2{t^3} - 8{t^2}} \right) - \left( {2{t^3} + {t^2} - 4t} \right) + \left( {4{t^2} + 2t - 8} \right) = 0\\
\Leftrightarrow 2{t^2}\left( {2{t^2} + t - 4} \right) - t\left( {2{t^2} + t - 4} \right) + 2.\left( {2{t^2} + t - 4} \right) = 0\\
\Leftrightarrow \left( {2{t^2} + t - 4} \right)\left( {2{t^2} - t + 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
2{t^2} + t - 4 = 0\\
2{t^2} - t + 2 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
t = \frac{{ - 1 + \sqrt {33} }}{4}\,\,\,\,\,\left( {t/m} \right)\\
t = \frac{{ - 1 - \sqrt {33} }}{4}\left( L \right)
\end{array} \right.\\
\Rightarrow t = \frac{{ - 1 + \sqrt {33} }}{4}\\
\Rightarrow x = {t^2} = \frac{{17 - \sqrt {33} }}{8}\\
y = 1 - x = \frac{{ - 9 + \sqrt {33} }}{8}
\end{array}\)
Vậy \(\left\{ \begin{array}{l}
x = \frac{{17 - \sqrt {33} }}{8}\\
y = \frac{{ - 9 + \sqrt {33} }}{8}
\end{array} \right.\)
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