Question Number 66472 by hmamarques1994@gmail.com last updated on 15/Aug/19
$$\begin{cases}{\sqrt[{\sqrt{\mathrm{6}}}]{\boldsymbol{\mathrm{x}}}+\sqrt[{\sqrt{\mathrm{5}}}]{\boldsymbol{\mathrm{y}}}=\mathrm{11}}\\{\frac{\sqrt[{\sqrt{\mathrm{5}}}]{\boldsymbol{\mathrm{y}}}}{\sqrt[{\sqrt{\mathrm{6}}}]{\boldsymbol{\mathrm{x}}}}=\mathrm{1}\frac{\mathrm{1}}{\mathrm{5}}}\end{cases} \\ $$$$\: \\ $$$$\:\boldsymbol{\mathrm{Qual}}\:\:\acute {\boldsymbol{\mathrm{e}}}\:\:\boldsymbol{\mathrm{o}}\:\:\boldsymbol{\mathrm{par}}\:\:\boldsymbol{\mathrm{ordenado}}\:\:\boldsymbol{\mathrm{na}}\:\:\boldsymbol{\mathrm{forma}}\:\:\boldsymbol{\mathrm{a}}^{\sqrt{\boldsymbol{\mathrm{p}}}} \:\:\boldsymbol{\mathrm{e}}\:\:\boldsymbol{\mathrm{b}}^{\sqrt{\boldsymbol{\mathrm{q}}}} \\ $$$$\:\boldsymbol{\mathrm{que}}\:\:\boldsymbol{\mathrm{satisfaz}}\:\:\boldsymbol{\mathrm{o}}\:\:\boldsymbol{\mathrm{sistema}}\:\:\boldsymbol{\mathrm{como}}\:\:\boldsymbol{\mathrm{possivel}}\:\:\boldsymbol{\mathrm{e}}\:\:\boldsymbol{\mathrm{determinado}}? \\ $$
Answered by MJS last updated on 15/Aug/19
$${p}=\sqrt[{\sqrt{\mathrm{6}}}]{{x}}={x}^{\frac{\sqrt{\mathrm{6}}}{\mathrm{6}}} \\ $$$$\mathrm{similar}\:{q}={y}^{\frac{\sqrt{\mathrm{5}}}{\mathrm{5}}} \\ $$$${p}+{q}=\mathrm{11} \\ $$$$\frac{{q}}{{p}}=\frac{\mathrm{6}}{\mathrm{5}} \\ $$$$\Rightarrow \\ $$$${p}=\mathrm{5} \\ $$$${q}=\mathrm{6} \\ $$$$\Rightarrow \\ $$$${x}=\mathrm{5}^{\sqrt{\mathrm{6}}} \approx\mathrm{51}.\mathrm{5371} \\ $$$${y}=\mathrm{6}^{\sqrt{\mathrm{5}}} \approx\mathrm{54}.\mathrm{9540} \\ $$
Commented by hmamarques1994@gmail.com last updated on 15/Aug/19
$$\boldsymbol{{Bom}}! \\ $$