Question

Efetue as operações.
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Answer 1

Resposta:

Explicação passo-a-passo:

$\frac{1}{\sqrt2}} + \frac{\sqrt{5}}{\sqrt{10}}=\frac{1.\sqrt{10} + 2.\sqrt{5}}{\sqrt{2}.\sqrt{10}}=\\\frac{\sqrt{10} + 2.\sqrt{5}}{\sqrt{2}.\sqrt{10}} = \frac{\sqrt{2.5} + 2.\sqrt{5}}{\sqrt{2}.\sqrt{2.5}} = \frac{\sqrt{2}.\sqrt{5} + 2.\sqrt{5}}{\sqrt{2}.\sqrt{2}.\sqrt{5}} = \frac{\sqrt{5}.(\sqrt{2}+2)}{\sqrt{2^2} .\sqrt{5}} = \frac{\sqrt{5}.(\sqrt{2} +2)}{2.\sqrt{5}} = \frac{\sqrt{2} + 2}{2}$

Letra b

$\frac{3}{2 - \sqrt{2}} - \frac{1}{\sqrt{2}+1} = \frac{3.(\sqrt{2}+1)-1.(2-\sqrt{2})}{(2-\sqrt{2}).(\sqrt{2}+1) }} = \frac{3.\sqrt{2} + 3-2+\sqrt{2}}{2.\sqrt{2}+2 -\sqrt{2}.\sqrt{2} -\sqrt{2}}= \\\frac{3\sqrt{2}+1+\sqrt{2}}{2\sqrt{2}+2-\sqrt{2^2}-\sqrt{2}} = \frac{4\sqrt{2} + 1 }{2\sqrt{2} + 2-2-\sqrt{2}} = \frac{4\sqrt{2}+1 }{2\sqrt{2}-\sqrt{2}}= \frac{4\sqrt{2}+1 }{\sqrt{2}} = \\\\\frac{(4\sqrt{2}+1).\sqrt{2}}{\sqrt{2}.\sqrt{2}} = \frac{4.\sqrt{2}.\sqrt{2} + \sqrt{2}}{\sqrt{2^2}} =$

$\frac{4.\sqrt{2^2} +\sqrt{2}}{2} = \frac{4.2+\sqrt{2} }{2}= \frac{8+\sqrt{2}}{2}$