Question

ajudaaa euu aquii pfv!!!!​

Answer 1

Explicação passo a passo:

$a)\\ {1.(3+\sqrt{6} )\over(3-\sqrt{6} )(3+\sqrt{6} )}=\\ \\ {3+\sqrt{6} \over3^2-(\sqrt{6} )^2}=\\ \\ {3+\sqrt{6} \over9-6}=\boxed{{3+\sqrt{6} \over3}}$

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$b)\\ {2(\sqrt{5} -\sqrt{3} )\over(\sqrt{5} +\sqrt{3} )(\sqrt{5} -\sqrt{3} )}=\\ \\ {2(\sqrt{5} -\sqrt{3} )\over(\sqrt{5})^2 -(\sqrt{3} )^2}=\\ \\ {2.(\sqrt{5} -\sqrt{3} )\over5-3}=\\ \\ {\not2.(\sqrt{5} -\sqrt{3} )\over\not2}=\boxed{\sqrt{5} -\sqrt{3} }$

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$c)\\{ (2-2\sqrt{2} )(2+\sqrt{2} )\over(2-\sqrt{2} )(2+\sqrt{2} )}=\\ \\ {2.2+2.\sqrt{2} -2\sqrt{2}.2-2.(\sqrt{2} )^2\over2^2-(\sqrt{2} )^2}=\\ \\ {\not4+2\sqrt{2} -4\sqrt{2} -\not4\over4-2}=\\ \\ {-\not2\sqrt{2} \over\not2}=\boxed{-\sqrt{2} }$

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$d)\\ {\sqrt{2} .(\sqrt{2} -1)\over(\sqrt{2} +1)(\sqrt{2} -1)}=\\ \\ {(\sqrt{2})^2 -1\sqrt{2} \over(\sqrt{2} )^2-1^1}=\\ \\ {2-\sqrt{2} \over2-1}={2-\sqrt{2} \over1}=\boxed{2-\sqrt{2} }$