Система уравнений x+y+(x+y)^1/2=20 и x^2+y^2=136

" alt="\left \{ {{x+y+(x+y)^1^/^2=20} \atop {x^2+y^2=136}} \right<=>" align="absmiddle" class="latex-formula">

" alt="\left \{ {{x+y+\sqrt{(x+y)}=20} \atop {x^2+y^2=136}} \right \ <=>" align="absmiddle" class="latex-formula">

Предлагаю заменить 0" alt="\sqrt{(x+y)}=p>0" align="absmiddle" class="latex-formula">

Тогда: $p^2+p=20; \ \ \ p^2+p-20=0; \ \ \ (p+5)(p-4)=0$

Получаем

p=- 5; p=4.

р>0 => p=4

Перейдем к начальной системе: \left \{ {{x+y}=16} \atop {x^2+y^2=136}} \right<=>" alt="\left \{ {{\sqrt{(x+y)}=4} \atop {x^2+y^2=136}} \right\ \ \ <=> \left \{ {{x+y}=16} \atop {x^2+y^2=136}} \right<=>" align="absmiddle" class="latex-formula">

$\left \{ {{x}=16-y} \atop {(16-y)^2+y^2=136}} \right$

Решим второе уравнение системы:

$16^2+y^2-32y+y^2=136$

$256+2y^2-32y-136=0$

$y^2-16y+60=0$

$(y-10)(y-6)=0$ - по Т.Виета

$y=10;\ \ \ \y=6$

Отсюда, подставляя получаем:

\left \{ {{x}=6} \atop {(y=10}} \right" alt=" \left \{ {{x}=16-y} \atop {y=10}} \right\ \ <=>\left \{ {{x}=6} \atop {(y=10}} \right" align="absmiddle" class="latex-formula">

\left \{ {{x}=10} \atop {(y=6}} \right" alt=" \left \{ {{x}=16-y} \atop {y=6}} \right\ \ <=>\left \{ {{x}=10} \atop {(y=6}} \right" align="absmiddle" class="latex-formula">

ОТВЕТ: (6;10); (10; 6)