Question

Resolva a equação:$( \frac{1}{ \sqrt{3} -1} + \frac{1}{ \sqrt{3}+1 } ) ^{-2}$

Answer 1

Oi

$\boxed{( \frac{1}{ \sqrt{3}-1 } + \frac{1}{ \sqrt{3}+1 } ) ^{-2} }$

$\boxed{( \frac{ \sqrt{3}+1 }{ \sqrt{3}^{2}-1 } + \frac{1}{ \sqrt{3}+1 } )^{-2} }$

$\boxed{( \frac{ \sqrt{3}+1 }{ 3-1 } + \frac{1}{ \sqrt{3}+1 } )^{-2} }$

$\boxed{( \frac{ \sqrt{3}+1 }{ 2 } + \frac{1}{ \sqrt{3}+1 } )^{-2} }$

$\boxed{( \frac{ \sqrt{3}+1 }{ 2 } + \frac{ \sqrt{3}-1 }{ \sqrt{3}^{2} -1 } )^{-2} }$

$\boxed{( \frac{ \sqrt{3}+1 }{ 2 } + \frac{ \sqrt{3}-1 }{ 3-1 } )^{-2} }$

$\boxed{( \frac{ \sqrt{3}+1 }{ 2 } + \frac{ \sqrt{3}-1 }{ 2 } )^{-2} }$

$\boxed{( \frac{ \sqrt{3}+1+\sqrt{3}-1 }{2}) ^{-2} }$

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

$\boxed{(\frac{2 \sqrt{3} }{2} )^{-2} = \sqrt{3} ^{-2} = 3^{- \frac{2}{2} } = 3^{-1} =\boxed{ \frac{1}{3} }}$