Question

308. Решите систему уравнений:
a) {x² +y²=40,
{ху=-12;​

Answer 1

$\displaystyle\bf\\\left \{ {{x^{2}+ y^{2} =40} \atop {xy=-12}} \right. \\\\\\\left \{ {{x^{2}+2xy+y^{2} =40-24 } \atop {xy=-12}} \right. \\\\\\\left \{ {{(x+y)^{2} =16} \atop {xy=-12}} \right.\\\\\\\left[\begin{array}{ccc}\left \{ {{x+y=-4} \atop {xy=-12}} \right. \\\left \{ {{x+y=4} \atop {xy=-12}} \right. \end{array}\right\\\\\\1)\\\\\left \{ {{x+y=-4} \atop {xy=-12}} \right. \\\\Teorema \ Vieta:$

$\displaystyle\bf\\\left[\begin{array}{ccc}\left \{ {{x=2} \atop {y=-6}} \right. \\\left \{ {{x=-6} \atop {y=2}} \right. \end{array}\right\\\\\\2)\\\\\left \{ {{x+y=4} \atop {xy=-12}} \right. \\\\\\\left[\begin{array}{ccc}\left \{ {{x=6} \atop {y=-2}} \right. \\\left \{ {{x=-2} \atop {y=6}} \right. \end{array}\right \\\\\\Otvet \ : \ (2 \ ; \ -6) \ , \ (-6 \ ; \ 2) \ , \ (6 \ ; \ -2) \ , \ (-2 \ ; \ 6)$

Answer 2

Ответ:вот

Объяснение: