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A melhor resposta!
2014-09-04T17:37:21-03:00
\boxed{f(x)=1- \frac{-3}{2x} }

a) f(-1/3) = substitui x na função por -1/3

f( \frac{-1}{3} )=1- \frac{3}{2* \frac{-1}{3} } }\\\\=1- \frac{3}{ \frac{-2}{3} } \\\\=1- \frac{3}{-2*3} \\\\1- \frac{-1}{2}\\\\\ = 1+\frac{1}{2} \\\\\ \boxed{ f( \frac{-1}{3}) = \frac{3}{2} }
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f (\sqrt{2}) =1- \frac{1}{2 \sqrt{2} } \\\\f( \sqrt{2})=  \boxed{\frac{2 \sqrt{2}-1 }{2 \sqrt{2} } }

ou vc poderia simplificar mais  racionalizando a raiz pra tirar do denominador 
\frac{(2 \sqrt{2}-1)* \sqrt{2}  }{(2 \sqrt{2})* \sqrt{2}  } \\\\= \frac{2*( \sqrt{2} )^2- \sqrt{2} }{2* (\sqrt{2})^2 }\\\\=   \frac{2*2- \sqrt{2} }{2*2} =\boxed{ \frac{4- \sqrt{2} }{4} }

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f(x)= \sqrt{ \frac{3}{2} } \\\\1- \frac{3}{2x}= \sqrt{ \frac{3}{2} }\\\\\\ \frac{2x-3}{2x}= \frac{ \sqrt{3} }{ \sqrt{2} }

racionalizando o lado direito
 \frac{ \sqrt{3} * \sqrt{2} }{ \sqrt{2} * \sqrt{2} }= \frac{ \sqrt{6} }{2}

agora temos 
] \frac{2x-3}{2x}=\frac{ \sqrt{6} }{2} \\\\2x-3= \frac{ \sqrt{6} }{2} *2x\\\\2x-3=x \sqrt{6} \\\\ -3=x \sqrt{6} -2x\\\\-3=x( \sqrt{6}-2) \\\\ \frac{-3}{( \sqrt{6} -2) }=x\\\\\ \boxed{\boxed{ \frac{3}{ 2- \sqrt{6}  }=x}}





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