Question

relacionar o denominador das seguintes trações:​

Answer 1

$\tt{a)}~\sf{\dfrac{1}{\sqrt{5}}=\dfrac{1}{\sqrt{5}}\cdot\dfrac{\sqrt{5}}{\sqrt{5}}}\\\sf{\dfrac{\sqrt{5}}{\sqrt{5^{\diagup\!\!\!\!2}}}=\dfrac{\sqrt{5}}{5}}\\\tt{b)} ~\sf{\dfrac{3}{\sqrt{3}}=\dfrac{3}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}}}\\\sf{\dfrac{3\sqrt{3}}{\sqrt{3^{\diagup\!\!\!\! 2}}}=\dfrac{ \diagup\!\!\!\! 3\cdot\sqrt{3}}{\diagup\!\!\!\!3}=\sqrt{3}}\\\tt{c)}~\sf{\dfrac{1}{2\sqrt{2}}=\dfrac{1}{2\sqrt{2}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}}}\\\sf{\dfrac{\sqrt{2}}{2\cdot\sqrt{2^{\diagup\!\!\!\!2}}}=\dfrac{\sqrt{2}}{4}}\\\tt{d)}~\sf{\dfrac{2}{\sqrt[3]{3}}=\dfrac{2}{\sqrt[3]{3}}\cdot\dfrac{\sqrt[3]{3^2}}{\sqrt[3]{3^2}}}\\\sf{\dfrac{2\sqrt[3]{9}}{\sqrt[\diagup\!\!\!\! 3]{3^{\diagup\!\!\!\!3}}}=\dfrac{2\sqrt[3]{9}}{3}}$