Racinalizando denominador da fraçao 5/6√16+6√196+³√49, obtemos ?
Katiakell:
é raiz cúbica de 49?
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Lembrar da relação:
![a^\frac{m}{n} \ \textless \ =\ \textgreater \ \sqrt[n]{a^m} a^\frac{m}{n} \ \textless \ =\ \textgreater \ \sqrt[n]{a^m}](https://tex.z-dn.net/?f=+a%5E%5Cfrac%7Bm%7D%7Bn%7D+%5C+%5Ctextless+%5C+%3D%5C+%5Ctextgreater+%5C+++%5Csqrt%5Bn%5D%7Ba%5Em%7D+)
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![\frac{5}{6}. \sqrt{16} + 6. \sqrt{196} + \sqrt[3]{49} \\ \\ \\ \frac{5}{6}. \sqrt{2^4} + 6. \sqrt{2^2 . 7^2} + 7^\frac{2}{3} \\ \\ \\ \frac{5}{6}. 2^2 + 6. 2 . 7 + 7^\frac{2}{3} \\ \\ \\ \frac{5}{6}. 4 + 84 + 7^\frac{2}{3} \\ \\ \\ \dfrac{5*4}{6} + 84 + \sqrt[3]{7}^2 \\ \\ \\ \dfrac{10}{3} + 84 + \sqrt[3]{7}^2 \\ \\ \\ =\ \textgreater \ \dfrac{3 .\sqrt[3]{7}^2 + 262}{3} \frac{5}{6}. \sqrt{16} + 6. \sqrt{196} + \sqrt[3]{49} \\ \\ \\ \frac{5}{6}. \sqrt{2^4} + 6. \sqrt{2^2 . 7^2} + 7^\frac{2}{3} \\ \\ \\ \frac{5}{6}. 2^2 + 6. 2 . 7 + 7^\frac{2}{3} \\ \\ \\ \frac{5}{6}. 4 + 84 + 7^\frac{2}{3} \\ \\ \\ \dfrac{5*4}{6} + 84 + \sqrt[3]{7}^2 \\ \\ \\ \dfrac{10}{3} + 84 + \sqrt[3]{7}^2 \\ \\ \\ =\ \textgreater \ \dfrac{3 .\sqrt[3]{7}^2 + 262}{3}](https://tex.z-dn.net/?f=+%5Cfrac%7B5%7D%7B6%7D.++%5Csqrt%7B16%7D+%2B+6.+%5Csqrt%7B196%7D++%2B++%5Csqrt%5B3%5D%7B49%7D++%5C%5C++%5C%5C++%5C%5C+%5Cfrac%7B5%7D%7B6%7D.++%5Csqrt%7B2%5E4%7D+%2B+6.+%5Csqrt%7B2%5E2+.+7%5E2%7D++%2B++7%5E%5Cfrac%7B2%7D%7B3%7D+%5C%5C++%5C%5C++%5C%5C+%5Cfrac%7B5%7D%7B6%7D.++2%5E2+%2B+6.+2+.+7++%2B++7%5E%5Cfrac%7B2%7D%7B3%7D++%5C%5C++%5C%5C++%5C%5C++%5Cfrac%7B5%7D%7B6%7D.++4+%2B+84++%2B++7%5E%5Cfrac%7B2%7D%7B3%7D++%5C%5C++%5C%5C++%5C%5C++%5Cdfrac%7B5%2A4%7D%7B6%7D+%2B+84+%2B++%5Csqrt%5B3%5D%7B7%7D%5E2++%5C%5C++%5C%5C++%5C%5C++%5Cdfrac%7B10%7D%7B3%7D++%2B+84+%2B++%5Csqrt%5B3%5D%7B7%7D%5E2++%5C%5C++%5C%5C++%5C%5C+%3D%5C+%5Ctextgreater+%5C+++%5Cdfrac%7B3+.%5Csqrt%5B3%5D%7B7%7D%5E2+%2B+262%7D%7B3%7D+)
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