Всё подходит
[tex]\displaystyle\left \{ {{9x^2+9y^2=13} \atop {3xy=2}} \right. \\\\\left \{ {{9x^2+9y^2=13} \atop {y=\dfrac{2}{3x} }} \right.[/tex]
Подставляем у в первое уравнение
[tex]\displaystyle\\9x^2+9(\frac{2}{3x})^2=13\\\\9x^2+9\times\frac{4}{9x^2}=13\\\\9x^2+\frac{4}{x^2} =13|\times x^2\neq0\\\\9x^4+4=13x^2\\9x^4-13x^2+4=0\\[/tex]
Заменяем переменную
[tex]x^2=t\\9t^2-13t+4=0\\D=(-13)^2-4\times9\times4=169-144=25=5^2\\\\t_{1,2}=\dfrac{13\pm5}{18} =\bigg[^\bigg{1}_\bigg{\dfrac{4}{9} }[/tex]
Возвращаем переменную
[tex]t=x^2\\\\x^2=1\\x_{1,2}=\pm1\\\\x^2=\dfrac{4}{9} \\\\x_{3,4}=\pm\dfrac{2}{3}[/tex]
Нашли иксы, теперь ищем игреки
[tex]\displaystyle y_1=\frac{2}{3\times1} =\frac{2}{3} \\\\y_2=\frac{2}{3\times(-1)} =-\frac{2}{3}\\\\y_3=\frac{2}{3\times\dfrac{2}{3} } =1\\\\y_4=\frac{2}{3\times(-\dfrac{2}{3}) } =-1[/tex]
Ответ: [tex]\displaystyle(1;\frac{2}{3}),(-1;-\frac{2}{3} ),(\frac{2}{3};1 ),(-\frac{2}{3} ;-1)[/tex]
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Answers & Comments
Всё подходит
[tex]\displaystyle\left \{ {{9x^2+9y^2=13} \atop {3xy=2}} \right. \\\\\left \{ {{9x^2+9y^2=13} \atop {y=\dfrac{2}{3x} }} \right.[/tex]
Подставляем у в первое уравнение
[tex]\displaystyle\\9x^2+9(\frac{2}{3x})^2=13\\\\9x^2+9\times\frac{4}{9x^2}=13\\\\9x^2+\frac{4}{x^2} =13|\times x^2\neq0\\\\9x^4+4=13x^2\\9x^4-13x^2+4=0\\[/tex]
Заменяем переменную
[tex]x^2=t\\9t^2-13t+4=0\\D=(-13)^2-4\times9\times4=169-144=25=5^2\\\\t_{1,2}=\dfrac{13\pm5}{18} =\bigg[^\bigg{1}_\bigg{\dfrac{4}{9} }[/tex]
Возвращаем переменную
[tex]t=x^2\\\\x^2=1\\x_{1,2}=\pm1\\\\x^2=\dfrac{4}{9} \\\\x_{3,4}=\pm\dfrac{2}{3}[/tex]
Нашли иксы, теперь ищем игреки
[tex]\displaystyle y_1=\frac{2}{3\times1} =\frac{2}{3} \\\\y_2=\frac{2}{3\times(-1)} =-\frac{2}{3}\\\\y_3=\frac{2}{3\times\dfrac{2}{3} } =1\\\\y_4=\frac{2}{3\times(-\dfrac{2}{3}) } =-1[/tex]
Ответ: [tex]\displaystyle(1;\frac{2}{3}),(-1;-\frac{2}{3} ),(\frac{2}{3};1 ),(-\frac{2}{3} ;-1)[/tex]