Question

Preciso dessa questão resolvida ainda hoje..

Se precisar da pergunta é:

 

 

Resolva as operações abaixo.

Answer 1

Seja $\text{x}=\dfrac{3}{2-\sqrt{2}}-\dfrac{1}{\sqrt{2}+1}$

 

Observe que:

 

$\dfrac{3}{2-\sqrt{2}}=\dfrac{3\cdot(2+\sqrt{2})}{2^2-(\sqrt{2})^2}=\dfrac{6+3\sqrt{2}}{2}$

 

E:

 

$\dfrac{1}{\sqrt{2}+1}=\dfrac{\sqrt{2}-1}{(\sqrt{2})^2-1^2}=\dfrac{\sqrt{2}-1}{1}=\sqrt{2}-1$

 

Logo:

 

$\text{x}=\dfrac{6+3\sqrt{2}}{2}-\sqrt{2}+1=\dfrac{6+3\sqrt{2}-2\sqrt{2}-2}{2}=\dfrac{4+\sqrt{2}}{2}$

Answer 2

$\frac{3}{2 - \sqrt{2}} - \frac{1}{\sqrt{2} + 1} = \\\\\\ \frac{3}{2 - \sqrt{2}} \times \frac{2 + \sqrt{2}}{2 + \sqrt{2}} - \frac{1}{\sqrt{2} + 1} \times \frac{\sqrt{2} - 1}{\sqrt{2} - 1} = \\\\\\ \frac{3(2 + \sqrt{2})}{(2 - \sqrt{2})(2 + \sqrt{2})} - \frac{1(\sqrt{2} - 1)}{(\sqrt{2} + 1)(\sqrt{2} - 1)} =$

 

$\frac{6 + 3\sqrt{2}}{4 - 2} - \frac{\sqrt{2} - 1}{2 - 1} = \\\\\\ \frac{6 + 3\sqrt{2}}{2_{/1}} - \frac{\sqrt{2} + 1}{1_{/2}} = \\\\\\ \frac{6 + 3\sqrt{2} - 2\sqrt{2} - 2}{2} = \\\\\\ \boxed{\frac{4 + \sqrt{2}}{2}}$