Question

{3√X+3√Y=5
{X+Y=35
Ps: выше кубические корни, это система, плиз с объяснением​

Answer 1

Решение задания прилагаю

Answer 2

$\left\{\begin{array}{l}\sqrt[3]{x}+\sqrt[3]{y}=5\\x+y=35\end{array}\right \\\\\sqrt[3]{x} =m \ ; \ \sqrt[3]{y}=n \ \Rightarrow \ x=m^{3} ; \ y=n^{3}\\\\\left\{\begin{array}{l}m+n=5\\m^{3}+n^{3}=35\end{array}\right\\\\\\\left\{\begin{array}{l}m+n=5\\(m+n)(m^{2}-mn+n^{2})=35\end{array}\right\\\\\left\{\begin{array}{l}m+n=5\\5\cdot(m^{2}-mn+n^{2})=35\end{array}\right\\\\\\\left\{\begin{array}{l}m+n=5\\m^{2}-mn+n^{2}=7\end{array}\right$

$\left\{\begin{array}{l}m+n=5\\(m+n)^{2}-3mn=7\end{array}\right\\\\\\\left\{\begin{array}{l}m+n=5\\5^{2} -3mn=7\end{array}\right\\\\\\\left\{\begin{array}{l}m+n=5\\25-3mn=7\end{array}\right\\\\\\\left\{\begin{array}{l}m+n=5\\3mn=18\end{array}\right\\\\\\\left\{\begin{array}{l}m+n=5\\mn=6\end{array}\right\\\\\\\left[\begin{array}{l}\left \{ {{m=2} \atop {n=3}} \right. \\\left \{ {{m=3} \atop {n=2}} \right. \end{array}\right$

$\left[\begin{array}{l}\left \{ {{\sqrt[3]{x}=2 } \atop {\sqrt[3]{y}=3 }} \right. \\\left \{ {{\sqrt[3]{x}=3 } \atop {\sqrt[3]{y}=2 }} \right. \end{array}\right\\\\\\\left[\begin{array}{ccc}\left \{ {{x=8} \atop {y=27}} \right. \\\left \{ {{x=27} \atop {y=8}} \right. \end{array}\right\\\\\\Otvet:\boxed{(8 \ ; \ 27) \ , \ (27 \ ; \ 8)}$