Question

Considerando x = (3^-1+6^-¹)/ ³√(1+9.16-¹) e y = (3^-2+2^-1)/ ∛(1-7.2^-3), os valores de x e y são respectivamente:

Answer 1

Na resolução foi muito usado:

• a transformação de potência com expoente negativo para fração : a^-b <=> a^(1/b)
• soma de frações, para isso foi deixado os denominadores iguais (consequentemente o numerador foi modificado também)
• a multiplicação dos meios e extremos : (a/b)/(c/d) <=> ad/bc
• racionalização: consiste em multiplicar a fração pelo denominador a fim de tranformar o denominador irracional em racional

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Descobrindo o valor de x:

$\begin{array}{l}\sf x=\dfrac{3^{-1}+6^{-1}}{\sqrt[\sf3]{\sf1+9\cdot16^{-1}}}\\\\\\\sf x=\dfrac{\dfrac{1}{3}+\dfrac{1}{6}}{\sqrt[\sf3]{\sf1+9\cdot\dfrac{1}{16}}}\\\\\\\sf x=\dfrac{\dfrac{2}{6}+\dfrac{1}{6}}{\sqrt[\sf3]{\sf1+\dfrac{9}{16}}}\\\\\\\sf x=\dfrac{\dfrac{2+1}{6}}{\sqrt[\sf3]{\sf\dfrac{1\cdot16+9}{16}}}\\\\\\\sf x=\dfrac{\dfrac{3}{6}}{\sqrt[\sf3]{\sf\dfrac{16+9}{16}}}\\\\\end{array}$

$\begin{array}{l}\sf x=\dfrac{\dfrac{1}{2}}{\sqrt[\sf3]{\sf\dfrac{25}{16}}}\\\\\\\sf x=\dfrac{\dfrac{1}{2}}{\dfrac{\sqrt[\sf3]{\sf\dfrac{25}{16}}}{1}}\\\\\sf x=\dfrac{1}{2\sqrt[\sf3]{\sf\dfrac{25}{16}}}\\\\\sf x=\dfrac{1}{\sf2\cdot\dfrac{\sqrt[\sf3]{\sf25}}{\sqrt[\sf3]{\sf16}}}\end{array}$

16 = 2⁴ = 2³ . 2, assim:

$\begin{array}{l} \sf x=\dfrac{1}{\sf2\cdot\dfrac{\sqrt[\sf3]{\sf25}}{\sqrt[\sf3]{\sf2^3\cdot2}}}\\\\\sf x=\dfrac{1}{\sf\diagdown\!\!\!\!2\cdot\dfrac{\sqrt[\sf3]{\sf25}}{\diagdown\!\!\!\!2\sqrt[\sf3]{\sf2}}}\\\\\sf x=\dfrac{1}{\sf\dfrac{\sqrt[\sf3]{\sf25}}{\sqrt[\sf3]{\sf2}}}\\\\\sf x=\dfrac{\dfrac{1}{1}}{\sf\dfrac{\sqrt[\sf3]{\sf25}}{\sqrt[\sf3]{\sf2}}}\\\\\sf x=\dfrac{\sqrt[\sf3]{\sf2}}{\sf\sqrt[\sf3]{\sf25}}\\\\\sf Racionalizando:\\\\\sf x=\dfrac{\sqrt[\sf3]{\sf2}}{\sf\sqrt[\sf3]{\sf5^2}}\\\\\sf x=\dfrac{\sqrt[\sf3]{\sf2}}{\sf\sqrt[\sf3]{\sf5^2}}\cdot\dfrac{\sqrt[\sf3]{\sf5}}{\sf\sqrt[\sf3]{\sf5}}\\\\\sf x=\dfrac{\sqrt[\sf3]{\sf10}}{\sf\sqrt[\sf3]{\sf5^{2+1}}}\\\\\sf x=\dfrac{\sqrt[\sf3]{\sf10}}{\sf\sqrt[\sf3]{\sf5^3}}\\\\\!\boxed{\sf x=\dfrac{\sqrt[\sf3]{\sf10}}{\sf 5}}\end{array}$

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Descobrindo o valor de y:

$\begin{array}{l}\sf y=\dfrac{3^{-2}+2^{-1}}{\sqrt[\sf3]{\sf1-7\cdot2^{-3}}}\\\\\\\sf y=\dfrac{\dfrac{1}{3^2}+\dfrac{1}{2}}{\sqrt[\sf3]{\sf1-7\cdot\dfrac{1}{2^3}}}\\\\\\\sf y=\dfrac{\dfrac{1}{9}+\dfrac{1}{2}}{\sqrt[\sf3]{\sf1-7\cdot\dfrac{1}{8}}}\\\\\\\sf y=\dfrac{\dfrac{2}{18}+\dfrac{9}{18}}{\sqrt[\sf3]{\sf1-\dfrac{7}{8}}}\\\\\end{array}$

$\begin{array}{l}\sf y=\dfrac{\dfrac{2+9}{18}}{\sqrt[\sf3]{\sf\dfrac{1\cdot8-7}{8}}}\\\\\\\sf y=\dfrac{\dfrac{11}{18}}{\sqrt[\sf3]{\sf\dfrac{1}{8}}}\\\\\\\sf y=\dfrac{\dfrac{11}{18}}{\dfrac{\sqrt[\sf3]{\sf\dfrac{1}{8}}}{1}}\\\\\\\sf y=\dfrac{11}{18\sqrt[\sf3]{\sf\dfrac{1}{8}}}\\\\\sf y=\dfrac{11}{18\cdot\dfrac{1}{2}}\\\\\sf y=\dfrac{11}{\dfrac{18}{2}}\\\\\!\boxed{\sf y=\dfrac{11}{9}}\\\\\end{array}$

Resposta: Assim, x = ³√10/5 e y = 11/9

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Att. Nasgovaskov