Question

Me ajudeem! Por favor, estou com dificuldades, como resolvo racionalização?

a) 6\ ∛12

b) 3\√₂

c) 30\√₁₂

d) 12√³

Answer 1

   a) $\frac{6}{ \sqrt[3]{12} } . \frac{ \sqrt[3]{12^2} }{ \sqrt[3]{12^2} }= \frac{6 \sqrt[3]{12^2} }{ \sqrt[3]{12.12^2} }= \frac{6 \sqrt[3]{12^2} }{ \sqrt[3]{12^3} }= \frac{6 \sqrt[3]{12^2} }{12} = \frac{ \sqrt[3]{12^2} }{2}$

b) $\frac{3}{ \sqrt{2} }. \frac{ \sqrt{2} }{ \sqrt{2} } = \frac{3 \sqrt{2} }{( \sqrt{2})^2 } = \frac{3 \sqrt{2} }{2}$

c) $\frac{30}{ \sqrt{12} } . \frac{ \sqrt{12} }{ \sqrt{12} }= \frac{30 \sqrt{12} }{( \sqrt{12})^2 } = \frac{30 \sqrt{12} }{12} = \frac{5 \sqrt{12} }{2}$

d)$\frac{12}{ \sqrt{3} }. \frac{ \sqrt{3} }{ \sqrt{3} } = \frac{12 \sqrt{3} }{( \sqrt{3})^2 } = \frac{12 \sqrt{3} }{3} =4 \sqrt{3}$