Simplifique a expressão
![(\sqrt[2]{x^2}(\sqrt[3]{x}(\sqrt[2]{x^4}))](/latexrender/pictures/1bbe11b4db364c71529e2a597a2a29ef.png "(\sqrt[2]{x^2}(\sqrt[3]{x}(\sqrt[2]{x^4}))")
, sendo x maior ou igual a 0, obtemos:espero q der pra entender que é uma raiz dentro da outra.
O que eu fiz foi transformar as raízes em potencias.
Ficando assim.
x^(2/2)*x^(1/3)*x^(4/2)
Eu cheguei em x^(10/3), Transformei em raiz de novo e ficou:
![\sqrt[3]{x^(10)}](/latexrender/pictures/862ee17653294eee3ed4f775863793a7.png "\sqrt[3]{x^(10)}")
.Passei o máximo de x pra fora e ficou
![{x}^{3}\sqrt[3]{x}](/latexrender/pictures/529daa6006354193bbceeefdacb5325d.png "{x}^{3}\sqrt[3]{x}")
.Infelizmente a resposta não é essa.
Seria
![x\sqrt[2]{x}](/latexrender/pictures/5c2f7feebca8e1954ca33463fad4deb6.png "x\sqrt[2]{x}")
.
![\\ \sqrt[2]{x^2\sqrt[3]{x\sqrt[2]{x^4}}} = \\\\ \sqrt[2]{x^2\sqrt[3]{x\cdot\,x^{\frac{4}{2}}}} = \\\\ \sqrt[2]{x^2\sqrt[3]{x \cdot\,x^2}}} = \\\\ \sqrt[2]{x^2\sqrt[3]{x^3}}} = \\\\ \sqrt[2]{x^2 \cdot x^{\frac{3}{3}}}} = \\\\ \sqrt[2]{x^2 \cdot x^1} = \\\\ \sqrt[2]{x^2} \cdot \sqrt[2]{x} = \\\\ x^{\frac{2}{2}} \cdot x^{\frac{1}{2}} = \\\\ \boxed{x \cdot \sqrt[2]{x}}](/latexrender/pictures/9d27f2c3881381478123acc634942226.png "\\ \sqrt[2]{x^2\sqrt[3]{x\sqrt[2]{x^4}}} = \\\\ \sqrt[2]{x^2\sqrt[3]{x\cdot\,x^{\frac{4}{2}}}} = \\\\ \sqrt[2]{x^2\sqrt[3]{x \cdot\,x^2}}} = \\\\ \sqrt[2]{x^2\sqrt[3]{x^3}}} = \\\\ \sqrt[2]{x^2 \cdot x^{\frac{3}{3}}}} = \\\\ \sqrt[2]{x^2 \cdot x^1} = \\\\ \sqrt[2]{x^2} \cdot \sqrt[2]{x} = \\\\ x^{\frac{2}{2}} \cdot x^{\frac{1}{2}} = \\\\ \boxed{x \cdot \sqrt[2]{x}}")
![\frac{\sqrt[]{\sqrt[4]{8}+\sqrt[]{\sqrt[]{2}-1}}-\sqrt[]{\sqrt[4]{8}-\sqrt[]{\sqrt[]{2}-1}}}{\sqrt[]{\sqrt[4]{8}-\sqrt[]{\sqrt[]{2}+1}}}](/latexrender/pictures/981987c7bcdf9f8f498ca4605785636a.png "\frac{\sqrt[]{\sqrt[4]{8}+\sqrt[]{\sqrt[]{2}-1}}-\sqrt[]{\sqrt[4]{8}-\sqrt[]{\sqrt[]{2}-1}}}{\sqrt[]{\sqrt[4]{8}-\sqrt[]{\sqrt[]{2}+1}}}")
![\sqrt[]{2}](/latexrender/pictures/f21662d1cabab6e8b273a4b6f1cd663a.png "\sqrt[]{2}")
e elevar ao quadrado os dois lados)