Предмет: Математика, автор: agqle

помогите пж
даю 50 баллов

Приложения:

Ответы

Автор ответа: kamilmatematik100504

Ответ $\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\la\la\la\la\ddddddddddddddddddddddddddddddddcleverdddddd\ffffffffffffffffffffffffffffffffffffffff\pppppppppppppppppppppppppppppppppppp\dddddd \displaystyle \large \boldsymbol{:}$

Пошаговое объяснение:

$1) \displaystyle \Large \boldsymbol{} \left \{ {{y-x=1} \atop {\!\!\!xy=6}} \right. => \left \{ {{y=x+1} \atop {x(x+1)=6}} \right. => \\\\\\x^2+x-6=0 \\\\\left \{ {{x_1+x_2=-1} \atop {\!\!\!\!x_1x_2=-6 }} \right. =>\left [ \ {{x_1=2} \atop {y_1=3}} \right. \ \ ; \ \ \left [\ {{x_2=-3} \atop {y_2=-2}} \right. \\\\\\OTBET : (2 \ ; \ 3) \ u \ (-3 \ \ ; \ \ - 2)$

$2) \displaystyle \Large \boldsymbol{} +\left \{ {{x^2+y=3} \atop {2x-y=-2}} \right. => \\\\\\ x^2+2x+y\!\!\!\!\diagup-y\!\!\!\!\diagup=3-2 \\\\x^2+2x-1=0 \\\\ D=4+4=8 \\\\x_{1;2 } =\frac{-2\pm2\sqrt{2} }{2} = -1 \pm\sqrt{2} \\\\\\ y_{1 }=2+2x_1 =2\sqrt{2} \\\\y_2=2+2x_2= -2\sqrt{2} \\\\\\ OTBET: (-1+\sqrt{2} \ \ ; \ \ 2\sqrt{2} ) \ \ u \ \ (-1-\sqrt{2} \ \ ; \ \ -2\sqrt{2} )$