Задача 32. Решите уравнение: \(2\,\left( {\,{x^4}-2{x^2} + 3} \right)\,\left( {{y^4}-3{y^2} + 4} \right) = 7.\)
Ответ
ОТВЕТ: \(\left( {\,1;\,\dfrac{{\sqrt 6 }}{2}} \right),\) \(\left( {\,1;\,-\dfrac{{\sqrt 6 }}{2}} \right),\;\;\;\left( {\,-1;\,\dfrac{{\sqrt 6 }}{2}} \right),\;\;\left( {\,-1;\,-\dfrac{{\sqrt 6 }}{2}} \right).\)
Решение
\(2\left( {{x^4}-2{x^2} + 3} \right)\left( {{y^4}-3{y^2} + 4} \right) = 7\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,2\left( {{{\left( {{x^2}-1} \right)}^2} + 2} \right)\left( {{{\left( {{y^2}-\dfrac{3}{2}} \right)}^2} + \dfrac{7}{4}} \right) = 7.\)
Так как \({\left( {{x^2}-1} \right)^2} + 2 \ge 2\) и \({\left( {{y^2}-\dfrac{3}{2}} \right)^2} + \dfrac{7}{4} \ge \dfrac{7}{4}\), то \(2\left( {{{\left( {{x^2}-1} \right)}^2} + 2} \right)\left( {{{\left( {{y^2}-\dfrac{3}{2}} \right)}^2} + \dfrac{7}{4}} \right) \ge 7.\)
Поэтому исходное уравнение будет иметь решение, если:
\(\left\{ \begin{array}{l}{\left( {{x^2}-1} \right)^2} + 2 = 2,\\{\left( {{y^2}-\dfrac{3}{2}} \right)^2} + \dfrac{7}{4} = \dfrac{7}{4}\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}{x^2}-1 = 0,\\{y^2}-\dfrac{3}{2} = 0\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}\left[ \begin{array}{l}x = 1,\\x = -1,\end{array} \right.\\\left[ \begin{array}{l}y = \dfrac{{\sqrt 6 }}{2},\\y = -\dfrac{{\sqrt 6 }}{2}.\end{array} \right.\end{array} \right.\)
Ответ: \(\left( {\,1;\,\dfrac{{\sqrt 6 }}{2}} \right),\) \(\left( {\,1;\,-\dfrac{{\sqrt 6 }}{2}} \right),\;\;\;\left( {\,-1;\,\dfrac{{\sqrt 6 }}{2}} \right),\;\;\left( {\,-1;\,-\dfrac{{\sqrt 6 }}{2}} \right).\)