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Resolva em C, o sistema
\( \Large\begin{cases}2\cdot log_{ \tfrac{1}{2} }(2y+x)=1,5\\log_{0,25}(5x-y)=-1\end{cases} \)


RESOLVENDO

Manipulando a primeira equação:
\( 2\cdot log_{(\frac{1}{2})}(2y+x)=1,5\\2\cdot log_{(2^{-1})}(x+2y)=\frac{3}{2}\\frac{2}{-1}\cdot log_{2}(x+2y)=\frac{3}{2}\\-2\cdot log_{2}(x+2y)=\frac{3}{2}\\log_{2}(x+2y)=-\frac{3}{4}\\2^{-\frac{3}{4}}=x+2y\\x+2y=(\frac{1}{2})^{\frac{3}{4}}\\x+2y=\sqrt[4]{\frac{1}{2^{3}}}\\x+2y=\sqrt[4]{\frac{1}{8}}\\x+2y=\frac{1}{\sqrt[4]{8}} \)
Racionalizando (multiplicar o numerador e denominador por raiz quarta de 8³):
\( x+2y=\frac{1\cdot\sqrt[4]{8^{3}}}{\sqrt[4]{8}\cdot\sqrt[4]{8^{3}}}\\x+2y=\frac{\sqrt[4]{512}}{8}\\x+2y=\frac{\sqrt[4]{2^{8}\cdot2}}{8}\\x+2y=\frac{2^{2}\sqrt[4]{2}}{8}\\boxed{\boxed{x+2y=\frac{\sqrt[4]{2}}{2}}} \)
Manipulando a segunda equação:
\( log_{(0,25)}(5x-y)=-1\\log_{(\frac{1}{4})}(5x-y)=-1\\log_{(2^{-2})}(5x-y)=-1\\frac{1}{-2}\cdot log_{2}(5x-y)=-1\\log_{2}(5x-y)=-(-2)\\log_{2}(5x-y)=2\\2^{2}=5x-y\\boxed{\boxed{5x-y=4}} \)
_________________
\( \begin{cases}x+2y=\dfrac{\sqrt[4]{2}}{2}\\5x-y=4\end{cases} \)
Multiplicando a segunda equação por 2:
\( \begin{cases}x+2y=\dfrac{\sqrt[4]{2}}{2}\\10x-2y=8\end{cases} \)
Somando membro à membro:
\( x+10x=\frac{\sqrt[4]{2}}{2}+8\\11x=\frac{\sqrt[4]{2}+2\cdot8}{2}\\boxed{\boxed{x=\frac{\sqrt[4]{2}+16}{22}}} \)
\( 5x-y=4\\y=5x-4\\y=\frac{5(\sqrt[4]{2}+16)}{22}-4\\y=\frac{5\sqrt[4]{2}+80}{22}-4\\y=\frac{5\sqrt[4]{2}+80-22\cdot4}{22}\\y=\frac{5\sqrt[4]{2}+80-88}{22}\\boxed{\boxed{y=\frac{5\sqrt[4]{2}-8}{22}}} \)



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