Giải hệ phương trình: \(\left\{ \begin{array}{l}{\left( {x + y} \right)^2} = xy + 3y - 1\\x + y = \dfrac{{{x^2} + y + 1}}{{1 + {x^2}}}\end{array} \right.\)
A.\(\left( {x;y} \right) = \left\{ {\left( {\dfrac{{ - 1 - \sqrt 5 }}{2};\dfrac{{5 + \sqrt 5 }}{2}} \right);\left( {\dfrac{{ - 1 + \sqrt 5 }}{2};\dfrac{{5 - \sqrt 5 }}{2}} \right)} \right\}\).
B.\(\left( {x;y} \right) = \left\{ {\left( {\dfrac{{ - 1 + \sqrt 5 }}{2};\dfrac{{5 + \sqrt 5 }}{2}} \right);\left( {\dfrac{{ - 1 - \sqrt 5 }}{2};\dfrac{{5 - \sqrt 5 }}{2}} \right)} \right\}\).
C.\(\left( {x;y} \right) = \left\{ {\left( {\dfrac{{ - 1 + \sqrt 5 }}{2};\dfrac{{-5 + \sqrt 5 }}{2}} \right);\left( {\dfrac{{ - 1 - \sqrt 5 }}{2};\dfrac{{-5 - \sqrt 5 }}{2}} \right)} \right\}\).
D.\(\left( {x;y} \right) = \left\{ {\left( {\dfrac{{ - 1 - \sqrt 5 }}{2};\dfrac{{-5 + \sqrt 5 }}{2}} \right);\left( {\dfrac{{ - 1 + \sqrt 5 }}{2};\dfrac{{-5 - \sqrt 5 }}{2}} \right)} \right\}\).
A.\(\left( {x;y} \right) = \left\{ {\left( {\dfrac{{ - 1 - \sqrt 5 }}{2};\dfrac{{5 + \sqrt 5 }}{2}} \right);\left( {\dfrac{{ - 1 + \sqrt 5 }}{2};\dfrac{{5 - \sqrt 5 }}{2}} \right)} \right\}\).
B.\(\left( {x;y} \right) = \left\{ {\left( {\dfrac{{ - 1 + \sqrt 5 }}{2};\dfrac{{5 + \sqrt 5 }}{2}} \right);\left( {\dfrac{{ - 1 - \sqrt 5 }}{2};\dfrac{{5 - \sqrt 5 }}{2}} \right)} \right\}\).
C.\(\left( {x;y} \right) = \left\{ {\left( {\dfrac{{ - 1 + \sqrt 5 }}{2};\dfrac{{-5 + \sqrt 5 }}{2}} \right);\left( {\dfrac{{ - 1 - \sqrt 5 }}{2};\dfrac{{-5 - \sqrt 5 }}{2}} \right)} \right\}\).
D.\(\left( {x;y} \right) = \left\{ {\left( {\dfrac{{ - 1 - \sqrt 5 }}{2};\dfrac{{-5 + \sqrt 5 }}{2}} \right);\left( {\dfrac{{ - 1 + \sqrt 5 }}{2};\dfrac{{-5 - \sqrt 5 }}{2}} \right)} \right\}\).
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