Question

Система уровнений 4x^2-2xy+y^2=4 _ 6x^2-3xy-y^2=-4

Answer 1

$+\left \{ {{4 x^{2} -2xy+ y^{2}=4 } \atop {6 x^{2} -3xy- y^{2} =-4}} \right.$
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10x² - 5xy = 0
2x² - xy = 0
2x(x - y) = 0
x(x - y ) = 0
x₁ = 0                 x - y = 0
                          x₂ =  y
Если x = 0 , то 4 * 0² - 2 * 0 * y + y² = 4
y² = 4    ⇒   y₁ = 2  ,    y₂ = - 2
Если x =  y , то 4 * ( y)² - 2 * ( y) * y + y² = 4
4y² - 2y² + y² = 4
3y² = 4
$y^{2} = \frac{4}{3} \\\\ y_{3}= \sqrt{ \frac{4}{3} }= \frac{2}{ \sqrt{3} }\\\\ y_{4} = -\sqrt{ \frac{4}{3} }=- \frac{2}{ \sqrt{3} } \\\\ x_{3}= \frac{2}{ \sqrt{3} }\\\\ x_{4} =- \frac{2}{ \sqrt{3} }$

Ответ:
$(0;2),(0;-2),( \frac{2}{ \sqrt{3} }; \frac{2}{ \sqrt{3} }),(- \frac{2}{ \sqrt{3} } ;- \frac{2}{ \sqrt{3} })$