Question

Simplifique ∛((x+1)²).√(2&(x+1)³)+3(x+1)√(6&(x+1))expressando numa base com expoente fracionario

Answer 1

$\sqrt[3]{(x+1)^{2}}\cdot \sqrt{(x+1)^{3}}+3(x+1)\cdot \sqrt[6]{x+1}$

$\sqrt[3]{(x+1)^{2}}\cdot \sqrt{(x+1)^{3}}+3(x+1)\cdot (x+1)^{\frac{1}{6}}\\ \\ \\ \sqrt[3\cdot 2]{(x+1)^{2\cdot 2}}\cdot \sqrt[2\cdot 3]{(x+1)^{3\cdot 3}}+3(x+1)^{1+\frac{1}{6}}\\ \\ \\ \sqrt[6]{(x+1)^{4}}\cdot \sqrt[6]{(x+1)^{9}}+3(x+1)^{\frac{6+1}{6}}\\ \\ \\ \sqrt[6]{(x+1)^{4}\cdot (x+1)^{9}}+3(x+1)^{\frac{7}{6}}\\ \\ \\ \sqrt[6]{(x+1)^{4+9}}+3(x+1)^{\frac{7}{6}}\\ \\ \\ \sqrt[6]{(x+1)^{13}}+3(x+1)^{\frac{7}{6}}\\ \\ \\ (x+1)^{\frac{13}{6}}+3(x+1)^{\frac{7}{6}}$

$(x+1)^{\frac{13}{6}}+3(x+1)^{\frac{7}{6}}\\ \\ \\ (x+1)^{\frac{7}{6}}\left(x+1+3\right)\\ \\ \\ (x+1)^{\frac{7}{6}}\left(x+4\right)$


Pode fazer dessa outra forma se achar melhor:

$\sqrt[3]{(x+1)^{2}}\cdot \sqrt{(x+1)^{3}}+3(x+1) \sqrt[6]{(x+1)}\\ \\ \\ (x+1)^{\frac{2}{3}}\cdot (x+1)^{\frac{3}{2}}+3(x+1)\cdot (x+1)^{\frac{1}{6}}\\ \\ \\ (x+1)^{\frac{2}{3}+\frac{3}{2}}+3(x+1)^{1+\frac{1}{6}}\\ \\ \\ (x+1)^{\frac{4+9}{6}}+3(x+1)^{\frac{6+1}{6}}\\ \\ \\ (x+1)^{\frac{13}{6}}+3(x+1)^{\frac{7}{6}}\\ \\ \\ (x+1)^{\frac{7}{6}}(x+1+3)\\ \\ \\ (x+1)^{\frac{7}{6}}(x+4)$