Question

A)__6__
√3
B)__2__
3√5
C)__4__
3-√2
Me ajudem nessa questão?

Answer 1

Olá.

$\mathsf{A}\\\mathsf{\dfrac{6}{\sqrt3}=\dfrac{6}{\sqrt3}\cdot\dfrac{\sqrt3}{\sqrt{3}}=\dfrac{6\sqr{3}}{3}}\\\\\\\\\mathsf{B}\\ \mathsf{\dfrac{2}{3\sqrt5}=\dfrac{2}{3\sqrt5}\cdot\dfrac{3\sqrt5}{3\sqrt5}=\dfrac{6\sqrt5}{9\cdot5}=\dfrac{6\sqrt5}{45}=\dfrac{2\sqrt5}{15}} \\\\\\\\\mathsf{C}\\\mathsf{\dfrac{4}{3-\sqrt2}=\dfrac{3+\sqrt2}{3+\sqrt2}=\dfrac{12+4\sqrt2}{9+3\sqrt2-3\sqrt2-2}=\dfrac{12+4\sqrt2}{7} }$

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Bons estudos.