Question

Решите системы уравнений.​

Answer 1

Решим систему методом подстановки.

$\left \{ {{x^{2}+y^{2} =10 } \atop {xy=-3}} \right.$

$\left \{ {{x^{2}+y^{2} =10 } \atop {x=-\frac{3}{y} }} \right.$

$\left \{ {{(-\frac{3}{y})^{2} +y^{2}=10 } \atop {x=-\frac{3}{y} }} \right.$

$\left \{ {{\frac{9}{y^{2} } +y^{2}=10 } \atop {x=-\frac{3}{y} }} \right.$

Выпишем первое уравнение отдельно и решим его.

$y^{2} *\frac{9}{y^{2} } +y^{2} *y^{2} =10*y^{2}$

$9+y^{4} =10y^{2}$

Решаем биквадратное уравнение:

$t=y^{2}$

$t^{2} -10t+9=0$

$D=(-10)^{2} -4*9=100-36=64$

$t_{1} =\frac{10+8}{2} =9$

$t_{2} =\frac{10-8}{2} =1$

$\left[\begin{array}{ccc}t_{1} =9\\t_{2} =1\end{array}\right$

$\left[\begin{array}{ccc}y^{2}=9 \\y^{2}=1 \end{array}\right$

$\left[\begin{array}{ccc}y=3\\y=-3\\y=1\\y=-1\end{array}\right$

Теперь найдем х:

$x=-\frac{3}{y}$

$x_{1} =\frac{-3}{3} =-1$

$x_{2} =\frac{-3}{-3} =1$

$x_{3} =\frac{-3}{1} =-3$

$x_{4} =\frac{-3}{-1} =3$

ответ: (-1;3), (1;-3), (-3;1), (3;-1)

Answer 2

$\left\{\begin{array}{ccc}x^{2}+y^{2} =10 \\xy=-3|\cdot2\end{array}\right\\\\\\+\left\{\begin{array}{ccc}x^{2}+y^{2} =10 \\2xy=-6\end{array}\right\\\\\\\left\{\begin{array}{ccc}x^{2}+2xy+y^{2} =4 \\2xy=-6\end{array}\right\\\\\\\left\{\begin{array}{ccc}(x+y)^{2}=4 \\xy=-3\end{array}\right\\\\\\1)\left\{\begin{array}{ccc}x+y=-2\\xy=-3\end{array}\right\\\\\\\left[\begin{array}{ccc}\left \{ {{x_{1} =-3} \atop {y_{1} =1}} \right. \\\left \{ {{x_{2}=1 } \atop {y_{2} =-3}} \right. \end{array}\right$

$2)\left\{\begin{array}{ccc}x+y=2 \\xy=-3\end{array}\right\\\\\\\left[\begin{array}{ccc}\left \{ {{x_{3}=3 } \atop {y_{3} =-1}} \right. \\\left \{ {{x_{4}=-1 } \atop {y_{4}=3 }} \right. \end{array}\right \\\\\\Otvet:\boxed{(-3 \ ; \ 1) \ , \ (1 \ ; \ -3) \ , \ (3 \ ; \ - 1) \ , \ (-1 \ ; \ 3)}$