Question

Me ajude por favoz presiso das respostas pra hoje.

Answer 1

$3 \sqrt{7} + 2 \sqrt{7} = 5 \sqrt{7} \\ 8 \sqrt{2} - 4 \sqrt{2} \: 3 \sqrt{2} =8\left(\sqrt{2}-3\right) \\ 4 \sqrt{10} - 9 \sqrt{10} = - 5 \sqrt{10} \\ \sqrt{2} + 3 \sqrt{5} - 2 \sqrt{2} + 4 \sqrt{5} = 7 \sqrt{5} \\ 4 \sqrt{7} - 3 \sqrt{5} - 2 \sqrt{2} + 4 \sqrt{5} - \sqrt{7} = 7\sqrt{5}-\sqrt{2}-\sqrt{7}$

$\sqrt{5} \times \sqrt{8} = 2 \sqrt{10} \\ \frac{ \sqrt[5]{30} }{ \sqrt[5]{15} } = \sqrt[5]{2} \\ \frac{ \sqrt[3]{160} }{ \sqrt[3]{5} } = 2^{\left(\frac{5}{3}\right)} \\ \frac{ \sqrt{96} }{ \sqrt{3} } = 4 \sqrt{2} \\ \sqrt[3]{2} \times \sqrt[3]{3} \times \sqrt[3]{5} = \sqrt[3]{30} \\ 3 \sqrt{5} \times 2 \sqrt{10} = 30 \sqrt{2} \\ \sqrt{2} \times \sqrt[3]{2} = 2^{\left(\frac{5}{6}\right)} \\ \sqrt[10]{ {3}^{7} } \div \sqrt{3} = \sqrt[5]{3}$

$\frac{2}{ \sqrt{10} } = \frac{ \sqrt{10} }{5} \\ \frac{6}{ \sqrt{6} } = \sqrt{6} \\ \frac{9}{ \sqrt{3} } = 3 \sqrt{3} \\ \frac{ \sqrt{5} }{2} = \frac{ \sqrt{5} }{2} \\ \frac{20}{2 \sqrt{5} } = 2 \sqrt{5} \\ \frac{2 \sqrt{5} }{5 \sqrt{2} } = \frac{ \sqrt{10} }{5}$

$\frac{1 - \sqrt{3} }{ \sqrt{3} } = \frac{ \sqrt{3} }{3} - 1 \\ \frac{3 - \sqrt{2} }{ \sqrt{2} } = \frac{3 \sqrt{2} }{2} - 1 \\ \frac{ \sqrt{5} + \sqrt{2} }{ \sqrt{5} } = \frac{\sqrt{2}+\sqrt{5}}{5} \\ \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} } = -\frac{\sqrt{6}}{3}+1 \\ \frac{2 + \sqrt{2} }{2} = \frac{\sqrt{2}}{2}+1 \\ \frac{1 + \sqrt{2} }{ \sqrt{5} } = \frac{\sqrt{5}\left(\sqrt{2}+1\right)}{5}$

$\frac{1}{3 - \sqrt{6} } = \frac{\sqrt{6}}{3}+1 \\ \frac{2}{ \sqrt{5} + \sqrt{3} } = \sqrt{5}-\sqrt{3} \\ \frac{2 - \sqrt{2} }{3 + \sqrt{2} } = \frac{8-5\sqrt{2}}{7} \\ \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } = 5-2\sqrt{6}$