Đáp án:
Vậy hệ phương trình có nghiệm \(\left( {x;\,\,y} \right) = \left\{ {\left( {\frac{{ - 45 + \sqrt {785} }}{{16}};\,\,\frac{{ - 25 + \sqrt {785} }}{{16}}} \right);\,\,\left( { - \frac{{45 + \sqrt {785} }}{{16}};\,\,\frac{{ - 25 - \sqrt {785} }}{{16}}} \right)} \right\}.\)
Giải thích các bước giải:
\(\begin{array}{l}
\left\{ \begin{array}{l}
x - 3y = \frac{{4y + x}}{y}\\
y - 3x = \frac{{9y + x}}{y}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x - 3y = \frac{{4y + x}}{y}\,\,\,\,\left( 1 \right)\\
x - y + 3\left( {x - y} \right) = - 5\,\,\,\,\,\left( 2 \right)
\end{array} \right.\,\,\,\,\left( {Dk:\,\,\,y \ne 0} \right)\\
\Rightarrow \left( 2 \right) \Leftrightarrow 4\left( {x - y} \right) = - 5\\
\Leftrightarrow x - y = - \frac{5}{4} \Leftrightarrow x = - \frac{5}{4} + y = \frac{{4y - 5}}{4}\\
\Rightarrow \left( 1 \right) \Leftrightarrow \frac{{4y - 5}}{4} - 3y = \frac{{4y + \frac{{4y - 5}}{4}}}{y}\\
\Leftrightarrow \frac{{4y - 5 - 12y}}{4} = \frac{{16y + 4y - 5}}{{4y}}\\
\Leftrightarrow - y\left( {5 + 8y} \right) = 20y - 5\\
\Leftrightarrow - 5y - 8{y^2} = 20y - 5\\
\Leftrightarrow 8{y^2} + 25y - 5 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
y = \frac{{ - 25 + \sqrt {785} }}{{16}}\,\,\,\left( {tm} \right) \Rightarrow x = \frac{{ - 45 + \sqrt {785} }}{{16}}\\
y = \frac{{ - 25 - \sqrt {785} }}{{16}}\left( {tm} \right) \Rightarrow x = - \frac{{45 + \sqrt {785} }}{{16}}
\end{array} \right..
\end{array}\)
Vậy hệ phương trình có nghiệm \(\left( {x;\,\,y} \right) = \left\{ {\left( {\frac{{ - 45 + \sqrt {785} }}{{16}};\,\,\frac{{ - 25 + \sqrt {785} }}{{16}}} \right);\,\,\left( { - \frac{{45 + \sqrt {785} }}{{16}};\,\,\frac{{ - 25 - \sqrt {785} }}{{16}}} \right)} \right\}.\)
