Задача 17. Решите систему уравнений \(\left\{ {\,\begin{array}{*{20}{c}} {\left( {x-y} \right)\,\left( {{x^2} + {y^2}} \right) = 65} \\ {\left( {x + y} \right)\,\left( {{x^2}-{y^2}} \right) = 5\,\,\,} \end{array}} \right.\)
ОТВЕТ: \(\left( {\,3;\,-2} \right),\;\left( {2;-3} \right)\).
Разделим первое уравнение на второе: \(\dfrac{{\left( {x-y} \right)\left( {{x^2} + {y^2}} \right)}}{{\left( {x + y} \right)\left( {x-y} \right)\left( {x + y} \right)}} = \dfrac{{65}}{5}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{{x^2} + {y^2}}}{{{x^2} + 2xy + {y^2}}} = 13\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\,13{x^2} + 26xy + 13{y^2} = {x^2} + {y^2}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,6{x^2} + 13xy + 6{y^2} = 0.\) Так как \(\left( {0;0} \right)\) не является решением исходной системы, то: \(6{x^2} + 13xy + 6{y^2} = 0\left| {:{y^2}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,6 \cdot {\left( {\dfrac{x}{y}} \right)^2} + 13 \cdot \frac{x}{y} + 6 = 0.\) Пусть \(\dfrac{x}{y} = t\), тогда: \(6{t^2} + 13t + 6 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = -\dfrac{3}{2},}\\{t = -\dfrac{2}{3}.}\end{array}} \right.\) \(\left[ {\begin{array}{*{20}{c}}{\dfrac{x}{y} = -\dfrac{3}{2},}\\{\dfrac{x}{y} = -\dfrac{2}{3}\,}\end{array}} \right.\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{y = -\dfrac{2}{3}x,}\\{y = -\dfrac{3}{2}x.}\end{array}} \right.\) Тогда исходная система равносильна совокупности двух систем: \(\left[ \begin{array}{l}\left\{ {\begin{array}{*{20}{c}}{y = -\dfrac{2}{3}x,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\left( {x + y} \right)\left( {{x^2}-{y^2}} \right) = 5,}\end{array}} \right.\\\left\{ {\begin{array}{*{20}{c}}{y = -\dfrac{3}{2}x,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\left( {x + y} \right)\left( {{x^2}-{y^2}} \right) = 5.}\end{array}} \right.\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\) Рассмотрим первую систему совокупности: \(\left\{ {\begin{array}{*{20}{c}}{y = -\dfrac{2}{3}x,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\left( {x + y} \right)\left( {{x^2}-{y^2}} \right) = 5}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{y = -\dfrac{2}{3}x,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\left( {x-\dfrac{2}{3}x} \right)\left( {{x^2}-\dfrac{4}{9}{x^2}} \right) = 5}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{y = -\dfrac{2}{3}x,}\\{{x^3} = 27\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x = 3,\,\,\,}\\{y = -2.}\end{array}} \right.\) Рассмотрим вторую систему совокупности: \(\left\{ {\begin{array}{*{20}{c}}{y = -\dfrac{3}{2}x,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\left( {x + y} \right)\left( {{x^2}-{y^2} = 5} \right)}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{y = -\dfrac{3}{2}x,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\left( {x-\dfrac{3}{2}x} \right)\left( {{x^2}-\dfrac{9}{4}{x^2}} \right) = 5}\end{array}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{y = -\dfrac{3}{2}x,}\\{{x^3} = 8\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 2,\,\,\,}\\{y = -3.}\end{array}} \right.} \right.\) Ответ: \(\left( {3;-2} \right),\,\,\,\left( {2;-3} \right).\)