Question

решите систему иррациональных уравнений

Answer 1

Ответ:

решение смотри на фотографии

Answer 2

$\displaystyle\bf\\\left \{ {{\sqrt[3]{x} \cdot\sqrt{y} =-12} \atop {\sqrt[3]{x} +\sqrt{y} =4}} \right. \\\\\sqrt[3]{x} =a \ \ ; \ \ \sqrt{y} =b \ , \ b\geq 0\\\\\\\left \{ {{a\cdot b=-12} \atop {a+b=4}} \right. \ - Teorema \ Vieta\\\\\\\left[\begin{array}{ccc}a=-2 \ ; \ b=6\\a=6 \ ; \ b=-2 < 0-neyd\end{array}\right\\\\\\\left \{ {{\sqrt[3]{x} =-2} \atop {\sqrt{y} =6}} \right. \\\\\\Otvet:\\\\\left \{ {{x=-8} \atop {y=36}} \right.$