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