Question

Como resolver essa potencia?

Answer 1

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

Answer 2

$\left(\frac{1}{\sqrt{2}}+\frac{\sqrt{2}}{2}\right)^2=2$

abrindo o quadrado da soma

$\left(\frac{1}{\sqrt{2}}+\frac{\sqrt{2}}{2}\right)^2=\left(\frac{1}{\sqrt{2}}\right)^2+2*\frac{1}{\sqrt{2}}*\frac{\sqrt{2}}{2}+\left(\frac{\sqrt{2}}{2}\right)^2$

$\left(\frac{1}{\sqrt{2}}+\frac{\sqrt{2}}{2}\right)^2=\frac{1}{2}+1+\frac{2}{4}$

$\left(\frac{1}{\sqrt{2}}+\frac{\sqrt{2}}{2}\right)^2=\frac{1}{2}+1+\frac{1}{2}$

$\left(\frac{1}{\sqrt{2}}+\frac{\sqrt{2}}{2}\right)^2=2$