Ответ:
(125;16)
Объяснение:
[tex]\displaystyle\bf\\\left \{ {{\sqrt[3]{x} -\sqrt[4]{y} =3} \atop {\sqrt[4]{y}\cdot\sqrt[3]{x} =10 } \right. \\\\\sqrt[3]{x} =m \ \ \ , \ \ \ \sqrt[4]{y} =n \ , \ n > 0\\\\\\\left \{ {{m-n=3} \atop {m\cdot n=10}} \right. \\\\\\\left \{ {{m=n+3} \atop {(n+3)\cdot n=10}} \right.\\\\\\\left \{ {{m=n+3} \atop {n^{2} +3n-10=0}} \right.\\\\\\\left \{ {{m=n+3} \atop {\left[\begin{array}{ccc}n_{1} =2\\n_{2} =-5-neyd\end{array}\right }} \right. \\\\\\\left \{ {{m=2+3=5} \atop {n=2}} \right.[/tex]
[tex]\displaystyle\bf\\\left \{ {{\sqrt[3]{x} =5} \atop {\sqrt[4]{y} =2}} \right. \ \ \ \Rightarrow \ \ \ \left \{ {{x=125} \atop {y=16}} \right. \\\\\\Otvet \ : \ \Big(125 \ ;\ 16\Big)[/tex]
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Answers & Comments
Ответ:
(125;16)
Объяснение:
[tex]\displaystyle\bf\\\left \{ {{\sqrt[3]{x} -\sqrt[4]{y} =3} \atop {\sqrt[4]{y}\cdot\sqrt[3]{x} =10 } \right. \\\\\sqrt[3]{x} =m \ \ \ , \ \ \ \sqrt[4]{y} =n \ , \ n > 0\\\\\\\left \{ {{m-n=3} \atop {m\cdot n=10}} \right. \\\\\\\left \{ {{m=n+3} \atop {(n+3)\cdot n=10}} \right.\\\\\\\left \{ {{m=n+3} \atop {n^{2} +3n-10=0}} \right.\\\\\\\left \{ {{m=n+3} \atop {\left[\begin{array}{ccc}n_{1} =2\\n_{2} =-5-neyd\end{array}\right }} \right. \\\\\\\left \{ {{m=2+3=5} \atop {n=2}} \right.[/tex]
[tex]\displaystyle\bf\\\left \{ {{\sqrt[3]{x} =5} \atop {\sqrt[4]{y} =2}} \right. \ \ \ \Rightarrow \ \ \ \left \{ {{x=125} \atop {y=16}} \right. \\\\\\Otvet \ : \ \Big(125 \ ;\ 16\Big)[/tex]