Cho hệ phương trình: \(\left\{ \begin{array}{l}2x - y = 1\\{x^2} + 2xy - {y^2} = 7\end{array} \right.\) , cặp nghiệm của hệ phương trình đã cho là

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\(\left( {x,\,\,y} \right) \in \left\{ {\left( {2\,;\,\,3} \right),\,\,\left( {4\,;\,\, - 9} \right)} \right\}\).

\(\left( {x,\,\,y} \right) \in \left\{ {\left( {2\,;\,\,3} \right),\,\,\left( { - 4\,;\,\, - 9} \right)} \right\}\).

\(\left( {x,\,\,y} \right) \in \left\{ {\left( {2\,;\,\, - 3} \right),\,\,\left( { - 4\,;\,\, - 9} \right)} \right\}\).

\[\left( {x,\,\,y} \right) \in \left\{ {\left( {2\,;\,\,3} \right),\,\,\left( {4\,;\,\,9} \right)} \right\}\].

Giải thích

Ta có\[\left\{ {\begin{array}{*{20}{l}}{2x - y = 1}\\{{x^2} + 2xy - {y^2} = 7}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{y = 2x - 1}\\{{x^2} + 2x\left( {2x - 1} \right) - {{\left( {2x - 1} \right)}^2} = 7}\end{array}} \right.\]

\[ \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{y = 2x - 1}\\{{x^2} + 4{x^2} - 2x - 4{x^2} + 4x - 1 - 7 = 0}\end{array}} \right.\]\[ \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{y = 2x - 1}\\{{x^2} + 2x - 8 = 0}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{y = 2x - 1}\\{\left( {x + 4} \right)\left( {x - 2} \right) = 0}\end{array}} \right.\]

\[ \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{y = 2x - 1}\\{\left[ {\begin{array}{*{20}{l}}{x + 4 = 0}\\{x - 2 = 0}\end{array}} \right.}\end{array}} \right. \Leftrightarrow \left\{ {\left\{ {\begin{array}{*{20}{l}}{y = 2x - 1}\\{\left[ {\begin{array}{*{20}{l}}{x = - 4}\\{x = 2}\end{array}} \right.}\end{array}} \right.} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{\left\{ {\begin{array}{*{20}{l}}{x = - 4}\\{y = - 9}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{l}}{x = 2}\\{y = 3}\end{array}} \right.}\end{array}} \right.\]. Chọn B.