Question

Помогите решить систему уравнений, Даю 50 баллов

Answer 1

Решение:

$\left \{ {{x^{2} +y^{2}=29 } \atop {y^{2}-4x^{2} =9 }} \right.$

x²+y²=29

y²=29-x²

29-x²-4x²=9

29-5x²=9

5x²=20

x²=4

x=±2

y²=29-2²        y²=29-(-2)²

y²=29-4          y²=29-4

y²=25             y²=25

y=±5               y=±5

Ответ: (2, 5)   (2,-5)   (-2, 5)   (-2, -5)  

Answer 2

Ответ:

Решение :

$x {}^{2} + y {}^{2} = 29 \\ y {}^{2} - 4x {}^{2} = 9 \\ \\ \\ y {}^{2} = 29 - x {}^{2} \\ y {}^{2} - 4x {}^{2} = 9 \\ \\ \\ 29 - x {}^{2} - 4x {}^{2} = 9 \\ \\ \\ x = - 2 \\ x = 2 \\ \\ \\ y {}^{2} = 29 - ( - 2) {}^{2} \\ y {}^{2} = 29 - 2 {}^{2} \\ \\ \\ y = - 5 \\ y = 5 \\ y = - 5 \\ y = 5 \\ \\ \\ (x1 \: \: \: y1) = ( - 2 \: \: \: - 5) \\ (x2 \: \: \: y2) = ( - 2 \: \: \: \: 5) \\( x3 \: \: \: y3) = (2 \: \: \: - 5) \\ (x4 \: \: \: y4) = (2 \: \: \: \: \: \: \: 5)$

Проверка :

$( - 2) {}^{2} + ( - 5) {}^{2} = 29 \\ ( - 5) {}^{2} - 4 \times ( - 2) {}^{2} = 9 \\ \\ \\ ( - 2) {}^{2} + 5 {}^{2} = 29 \\ 5 {}^{2} - 4 \times ( - 2) {}^{2} = 9 \\ \\ \\ 2 {}^{2} + ( - 5) {}^{2} = 29 \\ ( - 5) {}^{2} - 4 \times 2 {}^{2} = 9 \\ \\ \\ 2 {}^{2} + 5 {}^{2} = 29 \\ 5 {}^{2} - 4 \times 2 {}^{2} = 9 \\ \\ \\ \\ \\ \\ 29 = 29 \\ 9 = 9 \\ \\ \\ 29 = 29 \\ 9 = 9 \\ \\ \\ 29 = 29 \\ 9 = 9 \\ \\ \\ 29 = 29 \\ 9 = 9$