Nghiệm của hệ phương trình \[\left\{ {\begin{array}{*{20}{l}}{2{x^2} - 5xy + 2{y^2} = 0}\\{2{x^2} - {y^2} = 7}\end{array}} \right.\] là

4/150

\(\left( { - 2\,;{\mkern 1mu} {\mkern 1mu} \,1} \right),{\mkern 1mu} {\mkern 1mu} \,\left( {1\,;{\mkern 1mu} {\mkern 1mu} \,{\mkern 1mu} {\mkern 1mu} 2} \right).\)

\(\left( {1\,;{\mkern 1mu} {\mkern 1mu} \,2} \right),{\mkern 1mu} {\mkern 1mu} \,\left( {1\,;{\mkern 1mu} {\mkern 1mu} \,{\mkern 1mu} - 2} \right).\)

\(\left( { - 1\,;{\mkern 1mu} {\mkern 1mu} \,{\mkern 1mu} {\mkern 1mu} 2} \right),{\mkern 1mu} {\mkern 1mu} \,\left( {1\,;{\mkern 1mu} {\mkern 1mu} \,{\mkern 1mu} {\mkern 1mu} - 2} \right).\)

\(\left( {1\,;{\mkern 1mu} {\mkern 1mu} \,2} \right),{\mkern 1mu} {\mkern 1mu} \,\left( { - 1\,;{\mkern 1mu} {\mkern 1mu} \, - 2} \right).\)

Giải thích

Ta có \(\left\{ {\begin{array}{*{20}{l}}{2{x^2} - 5xy + 2{y^2} = 0{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \left( 1 \right)}\\{2{x^2} - {y^2} = 7{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \left( 2 \right)}\end{array}} \right.\)

Khi đó\(\left( 1 \right) \Leftrightarrow \left( {2x - y} \right)\left( {x - 2y} \right) = 0 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{y = 2x}\\{x = 2y}\end{array}} \right.\)

•Với \(y = 2x\), ta có: \(\left( 2 \right) \Leftrightarrow 2{x^2} - {\left( {2x} \right)^2} = 7 \Leftrightarrow - 2{x^2} = 7{\mkern 1mu} {\mkern 1mu} \left( {ktm} \right)\)

•Với \(x = 2y\), ta có: \(\left( 2 \right) \Leftrightarrow 2.{\left( {2y} \right)^2} - {y^2} = 7 \Leftrightarrow 7{y^2} = 7 \Leftrightarrow y = \pm 1\)\( \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{y = 1 \Rightarrow x = 2}\\{y = - 1 \Rightarrow x = - 2}\end{array}} \right..\)

Vậy hệ phương trình đã cho có hai nghiệm\(\left( {1\,;\,{\mkern 1mu} {\mkern 1mu} 2} \right)\) và \(\left( { - 1\,;\,{\mkern 1mu} {\mkern 1mu} - 2} \right).\) Chọn D.