Question

Поможіть пж терміново люди

Answer 1

$\displaystyle\bf\\1)\\\\+\left \{ {{y+x^{2} =10} \atop {y-x^{2} =-8}} \right. \\---------\\2y=2\\\\y=1\\\\x^{2} =10-y=10-1=9\\\\x_{1} =-\sqrt{9} =-3 \ \ ; \ \ x_{2} =\sqrt{9} =3\\\\\\Otvet \ : \ (-3 \ ; \ 1) \ \ , \ \ (3 \ ; \ 1)\\\\\\2)\\\\+\left \{ {{x^{2} +y=1} \atop {-x^{2} +4y=19}} \right. \\---------\\5y=20\\\\y=4\\\\x^{2} =1-y=1-4=-3$

Ответ : решений нет

$\displaystyle\bf\\1)\\\\+\left \{ {{x+y^{2} =3} \atop {x-y^{2} =1}} \right. \\---------\\2x=4\\\\x=2\\\\y^{2} =3-x=3-2=1\\\\y_{1} =-\sqrt{1} =-1 \ \ ; \ \ y_{2} =\sqrt{1} =1\\\\\\Otvet \ : \ (2 \ ; \ -1) \ \ , \ \ (2 \ ; \ 1)$