a) 4 b) 6 xó 8 d) 12 6- Calcule: a) sqrt (25)+sqrt [3](27)+sqrt [4](81)= b) sqrt (64)+sqrt [3](-64)+sqrt [6](64)= 6- Reduza a um único radical: a) sqrt (sqrt (10))= sqrt (sqrt (sqrt {2))}= c) sqrt [3](sqrt (3))= sqrt [3](sqrt [3](sqrt {3))}= então o valor de x+y então o valor de x+yé:de 81 dividido pela raiz cúbica de 2 x=(sqrt [5](32))/(sqrt (4)) e y=(sqrt [4](81))/(sqrt [3](27))

Vamos corrigir e detalhar as respostas:<br /><br />6- Calcule:<br />a) \(\sqrt{25} + \sqrt[3]{27} + \sqrt[4]{81}\)<br /> \[<br /> \sqrt{25} = 5, \quad \sqrt[3]{27} = 3, \quad \sqrt[4]{81} = 3<br /> \]<br /> \[<br /> \sqrt{25} + \sqrt[3]{27} + \sqrt[4]{81} = 5 + 3 + 3 = 11<br /> \]<br /><br />b) \(\sqrt{64} + \sqrt[3]{-64} + \sqrt[6]{64}\)<br /> \[<br /> \sqrt{64} = 8, \quad \sqrt[3]{-64} = -4, \quad \sqrt[6]{64} = 2<br /> \]<br /> \[<br /> \sqrt{64} + \sqrt[3]{-64} + \sqrt[6]{64} = 8 + (-4) + 2 = 6<br /> \]<br /><br />6- Reduza a um único radical:<br />a) \(\sqrt{\sqrt{10}}\)<br /> \[<br /> \sqrt{\sqrt{10}} = \sqrt[4]{10}<br /> \]<br /><br />b) \(\sqrt{\sqrt{\sqrt{2}}}\)<br /> \[<br /> \sqrt{\sqrt{\sqrt{2}}} = \sqrt[8]{2}<br /> \]<br /><br />c) \(\sqrt[3]{\sqrt{3}}\)<br /> \[<br /> \sqrt[3]{\sqrt{3}} = \sqrt[6]{3}<br /> \]<br /><br />d) \(\sqrt[3]{\sqrt[3]{\sqrt{3}}}\)<br /> \[<br /> \sqrt[3]{\sqrt[3]{\sqrt{3}}} = \sqrt[9]{3}<br /> \]<br /><br />7. \(x = \text{raiz quinta de 32 dividido pela raiz quadrada de 4}\) e \(y = \text{raiz}\)<br /> \[<br /> x = \frac{\sqrt[5]{32}}{\sqrt{4}}, \quad y = \sqrt[4]{81}<br /> \]<br /> \[<br /> \sqrt[5]{32} = 2, \quad \sqrt{4} = 2, \quad \sqrt[4]{81} = 3<br /> \]<br /> \[<br /> x = \frac{2}{2} = 1, \quad y = 3<br /> \]<br /> \[<br /> x + y = 1 + 3 = 4<br /> \]<br /><br />Portanto, as respostas corrigidas são:<br />6a) 11<br />6b) 6<br />6c) \(\sqrt[4]{10}\)<br />6d) \(\sqrt[8]{2}\)<br />6e) \(\sqrt[6]{3}\)<br />6f) \(\sqrt[9]{3}\)<br />7) \(x + y = 4\)