Question

racionalize o denominador e simplifique quando for possível.

$a) \frac{4}{3 \sqrt{2} }$

$b) \frac{ \sqrt{7} }{ \sqrt{6} }$

$c) \frac{ \sqrt{3}{+} {\sqrt{5} } }{ \sqrt{3} }$

$d) \frac{ \sqrt{2}{+1} }{ \sqrt{2} }$

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Answer 1

Explicação passo-a-passo:

$a)\\ {4\over3\sqrt{2} }={4\sqrt{2} \over3.\sqrt{2} .\sqrt{2} }={4\sqrt{2} \over3.\sqrt{4} }={4\sqrt{2} \over3.2}={4\sqrt{2} \over6}={2\sqrt{2} \over3}\\ \\ b)\\ {\sqrt{7} \over\sqrt{6} }={\sqrt{7}.\sqrt{6}  \over\sqrt{6} .\sqrt{6} }={\sqrt{42} \over\sqrt{36} }={\sqrt{42} \over6}$

$c)\\ {\sqrt{3} +\sqrt{5} \over\sqrt{3} }={\sqrt{3} .(\sqrt{3} +\sqrt{5} )\over\sqrt{3} .\sqrt{3} }={\sqrt{9} +\sqrt{15} \over\sqrt{9} }={3+\sqrt{15} \over3}\\ \\ d)\\ {\sqrt{2} .(\sqrt{2} +1)\over\sqrt{2} .\sqrt{2} }={\sqrt{4} +\sqrt{2} \over\sqrt{4} }={2+\sqrt{2} \over2}$