Question

ДЕДЛАЙН ЧЕРЕЗ ЧАС :(
Решите пожалуйста 3 системы уравнений.
Подробно.

Answer 1

Відповідь:

Покрокове пояснення:

1. $\left \{ {{x + y=5} \atop {x^{2} + y^{2} =17}} \right.$

$\left \{ {{(x + y)^{2} =25} \atop {x^{2} + y^{2} =17}} \right.$

$\left \{ {{x^{2} + 2xy + y^{2} =25} \atop {x^{2} + y^{2} =17}} \right.$

$\left \{ {{2xy=8} \atop {x + y = 5}} \right.$

$\left \{ {{2xy=8} \atop {x=5-y }} \right.$

$\left \{ {{2(5-y)y=8} \atop {x=5 - y }} \right.$

$\left \{ {{10y - 2y^{2} =8} \atop {x=5 - y}} \right.$

$\left \{ {{y^{2} - 5y + 4 =0} \atop {x=5-y}} \right.$

$\left \{ {{y_{1} =1, y_{2} =4} \atop {x=5-y}} \right.$

$x_{1} = 5 - 1 =4$

$x_{2} = 5 - 4 = 1$

(4; 1), (1; 4)

2.

$\left \{ {{x - y =2} \atop {x^{2} - y^{2} =8 }} \right.$

$\left \{ {{x - y=2} \atop {(x+y)(x-y)=8}} \right.$

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

$2x = 6$

$x = 3$

$y = 4 - x = 4 - 3 = 1$

(3; 1)

3.

$\left \{ {{x^{2} + xy - y^{2} =11} \atop {x-2y=1}} \right.$

$\left \{ {{x^{2} + xy - y^{2} =11} \atop {x=1 + 2y}} \right.$

$\left \{ {{(1+2y)^{2} + (1 + 2y)y - y^{2} =11} \atop {x=1 + 2y}} \right.$

$\left \{ {{1 + 4y + 4y^{2} + y + 2y^{2} - y^{2} =11} \atop {x=1 + 2y}} \right.$

$\left \{ {{5y^{2} + 5y - 10 = 0} \atop {x=1 + 2y}} \right.$

$\left \{ {{y^{2} + y - 2 = 0} \atop {x=1 + 2y}} \right.$

$\left \{ y_{1} = -2, y_{2} = 1 }\atop {x=1 + 2y}} \right.$

$x_{1} = 1 + 2y_{1} = 1 - 4 = -3$

$x_{2} = 1 + 2y_{2} = 1 + 2 = 3$

(-3; -2), (3; 1)