Giải hệ phương trình :
\(\left\{ \begin{array}{l}{x^2} + {y^2} + 6xy - \frac{1}{{{{\left( {x - y} \right)}^2}}} + \frac{9}{8} = 0\\2y - \frac{1}{{x - y}} + \frac{5}{4} = 0\end{array} \right.\)
A.\(\left( {x;y} \right) = \left( {\frac{-7}{8};\frac{-3}{8}} \right),\,\,\left( {\frac{{13}}{8}; - \frac{3}{8}} \right)\)
B.\(\left( {x;y} \right) = \left( {\frac{-7}{8};\frac{3}{8}} \right),\,\,\left( {\frac{{-13}}{8}; - \frac{3}{8}} \right)\)
C.\(\left( {x;y} \right) = \left( {\frac{7}{8};\frac{3}{8}} \right),\,\,\left( {\frac{{13}}{8}; - \frac{3}{8}} \right)\)
D.\(\left( {x;y} \right) = \left( {\frac{7}{8};\frac{3}{8}} \right),\,\,\left( {\frac{{-13}}{8}; \frac{3}{8}} \right)\)
\(\left\{ \begin{array}{l}{x^2} + {y^2} + 6xy - \frac{1}{{{{\left( {x - y} \right)}^2}}} + \frac{9}{8} = 0\\2y - \frac{1}{{x - y}} + \frac{5}{4} = 0\end{array} \right.\)
A.\(\left( {x;y} \right) = \left( {\frac{-7}{8};\frac{-3}{8}} \right),\,\,\left( {\frac{{13}}{8}; - \frac{3}{8}} \right)\)
B.\(\left( {x;y} \right) = \left( {\frac{-7}{8};\frac{3}{8}} \right),\,\,\left( {\frac{{-13}}{8}; - \frac{3}{8}} \right)\)
C.\(\left( {x;y} \right) = \left( {\frac{7}{8};\frac{3}{8}} \right),\,\,\left( {\frac{{13}}{8}; - \frac{3}{8}} \right)\)
D.\(\left( {x;y} \right) = \left( {\frac{7}{8};\frac{3}{8}} \right),\,\,\left( {\frac{{-13}}{8}; \frac{3}{8}} \right)\)
Loga
Hóa Học lớp 12
