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Автор - ЭмилиСтоун11

HELP Пожалуйста, помогите решить

Ответ

Автор - nikebod313

left { {bigg{dfrac{x}{y} - dfrac{y}{x} = dfrac{3}{2}    } atop bigg{x^{2} + y^{2} = 45}} right.

Пусть dfrac{x}{y} = t. Тогда dfrac{y}{x} = dfrac{1}{dfrac{x}{y} } = dfrac{1}{t} и первое уравнение примет вид:

t - dfrac{1}{t} = dfrac{3}{2}

Решим уравнение относительно t neq 0

dfrac{2t^{2} - 3t - 2}{t} = 0\2t^{2} - 3t - 2 = 0\t_{1} = -dfrac{1}{2} \t_{2} = 2

Обратная замена:

left[begin{array}{ccc}left { {bigg{dfrac{x}{y} = -dfrac{1}{2}       } atop bigg{x^{2} + y^{2} = 45}}\ left { {bigg{dfrac{x}{y} = 2         } atop bigg{x^{2} + y^{2} = 45}} \ end{array}right

left[begin{array}{ccc}left { {bigg{x = -dfrac{y}{2}        } atop bigg{x^{2} + y^{2} = 45}}\ left { {bigg{x = 2y         } atop bigg{x^{2} + y^{2} = 45}} \ end{array}right

left[begin{array}{ccc}left { {bigg{x = -dfrac{y}{2}               } atop bigg{ bigg(-dfrac{y}{2} bigg)^{2} + y^{2} = 45}}\ left { {bigg{x = 2y           } atop bigg{ (2y)^{2} + y^{2} = 45}} \ end{array}right

left[begin{array}{ccc}left { {bigg{x = -dfrac{y}{2}         } atop bigg{ dfrac{y^{2}}{4} + y^{2} = 45}}\ left { {bigg{x = 2y          } atop bigg{4y^{2} + y^{2} = 45}} \ end{array}right

left[begin{array}{ccc}left { {bigg{x = -dfrac{y}{2}   } atop bigg{ dfrac{5y^{2}}{4} = 45}}\ left { {bigg{x = 2y   } atop bigg{5y^{2} = 45}} \ end{array}right

left[begin{array}{ccc}left { {bigg{x = -dfrac{y}{2} } atop bigg{ y^{2} = 36}}\ left { {bigg{x = 2y } atop bigg{y^{2} = 9}} \ end{array}right

left[begin{array}{ccc}left { {bigg{x = -dfrac{y}{2} } atop bigg{ y = pm 6}}\ left { {bigg{x = 2y } atop bigg{y = pm 3}} \ end{array}right

Если y = 6, то x = -dfrac{6}{2} = -3

Если y = -6, то x = -dfrac{-6}{2} = 3

Если y = 3, то x = 2 cdot 3 = 6

Если y = -3, то x = 2 cdot (-3) = -6

Ответ: (-3; 6),  (3; -6),  (6; 3),  (-6; - 3).

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