Задача 12. Решите систему уравнений \(\left\{ {\,\begin{array}{*{20}{c}} {{x^2}-3xy + 2{y^2} = 3\,\,\,} \\ {2{x^2}-2xy-{y^2} = -6} \end{array}} \right.\)
ОТВЕТ: \(\left( {\,1;\,2} \right),\;\left( {\,-1;\,-2} \right)\).
\(\left\{ {\begin{array}{*{20}{c}}{{x^2}-3xy + 2{y^2} = 3\left| { \cdot 2,} \right.}\\{2{x^2}-2xy-{y^2} = -6\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{2{x^2}-6xy + 4{y^2} = 6,}\\{2{x^2}-2xy-{y^2} = -6.}\end{array}} \right.\) Прибавим к первому уравнению второе: \(4{x^2}-8xy + 3{y^2} = 0.\) Так как \(\left( {0;0} \right)\) не является решением исходной системы, то: \(4{x^2}-8xy + 3y = 0\left| {:{y^2}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,4{\left( {\dfrac{x}{y}} \right)^2}-8\dfrac{x}{y} + 3 = 0.\) Пусть \(\dfrac{x}{y} = t\), тогда: \(4{t^2}-8t + 3 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = \dfrac{3}{2},}\\{t = \dfrac{1}{2}.\,}\end{array}} \right.\) \(\left[ {\begin{array}{*{20}{c}}{\dfrac{x}{y} = \dfrac{3}{2},}\\{\dfrac{x}{y} = \dfrac{1}{2}\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{y = \dfrac{2}{3}x,}\\{y = 2x.}\end{array}} \right.\) Тогда исходная система равносильна совокупности двух систем: \(\left[ \begin{array}{l}\left\{ {\begin{array}{*{20}{c}}{y = \dfrac{2}{3}x,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{x^2}-3xy + 2{y^2} = 3}\end{array}} \right.\\\left\{ {\begin{array}{*{20}{c}}{y = 2x,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{x^2}-3xy + 2{y^2} = 3.\,}\end{array}} \right.\end{array} \right.\) Рассмотрим первую систему совокупности: \(\left\{ {\begin{array}{*{20}{c}}{y = \dfrac{2}{3}x,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{x^2}-3xy + 2{y^2} = 3\,\,\,\,}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{y = \dfrac{2}{3}x,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{x^2}-2{x^2} + \dfrac{8}{9}{x^2} = 3\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{y = \dfrac{2}{3}x,\,\,\,}\\{{x^2} = -27\,}\end{array}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\emptyset .} \right.\) Рассмотрим вторую систему совокупности: \(\left\{ {\begin{array}{*{20}{c}}{y = 2x,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{x^2}-3xy + 2{y^2} = 3}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{y = 2x,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{x^2}-6{x^2} + 8{x^2} = 3\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{y = 2x,}\\{{x^2} = 1\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{y = 2x,}\\{\left[ {\begin{array}{*{20}{c}}{x = 1,\,\,}\\{x = -1}\end{array}} \right.}\end{array}} \right.\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{x = 1,}\\{y = 2,}\end{array}\,\,\,} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{x = -1,}\\{y = -2.}\end{array}} \right.}\end{array}} \right.\) Ответ: \(\left( {1;2} \right),\,\,\,\left( {-1;-2} \right).\)