Автор - Schizik

Решите уравнение, пожалуйста) ​

Ответ

Проверено экспертом

Автор - Universalka

left { {{sqrt[3]{x}+sqrt[3]{y}=4} atop {x+y=28}} right. \\sqrt[3]{x}=m;sqrt[3]{y}=n\\left { {{m+n=4} atop {m^{3}+n^{3}=28}} right. \\:left { {{(m+n)(m^{2}-mn+n^{2})=28} atop {m+n=4}} right.\-----------\m^{2}-mn+n^{2}=7\\n=4-m\\m^{2} -m*(4-m) +(4-m)^{2}=7\\m^{2}-4m+m^{2}+16-8m+m^{2}-7=0\\3m^{2}-12m+9=0\\m^{2}-4m+3=0\\m_{1}=1\\m_{2}=3\\n_{1}=4-1=3\\n_{2}=4-3=1

1)left { {{sqrt[3]{x} =1} atop {sqrt[3]{y} =3}} right.\\left { {{x=1} atop {y=27}} right.\\2)left { {{sqrt[3]{x} =3} atop {sqrt[3]{y}=1 }} right.\\left { {{y=x=27} atop {y=1}} right. \\Otvet:boxed{(27;1),(1;27)}

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