Question

Resolva as expressões com radicais abaixo.
POR FAVOR!Me ajudem!Não consigo fazer nenhuma!

Resposta gabarito:
20a)3
b)raiz de 2
c)2+2raiz de 2

Answer 1

a)
$\frac{1}{3+ \sqrt{7} } + \frac{1}{3- \sqrt{7} } = \\ \\ \frac{3- \sqrt{7}+3+ \sqrt{7} }{(3+ \sqrt{7})(3- \sqrt{7} } = \\ \\ \frac{6}{3^2- \sqrt{7^2} } = \\ \\ \frac{6}{9-7} = \frac{6}{2} =\fbox{ 3 }$


b)
$\sqrt{3}. \sqrt{6} - \frac{4}{ \sqrt{2} } = \\ \\ \sqrt{18} - \frac{4}{ \sqrt{2} } = \\ \\ \frac{ \sqrt{2}. \sqrt{18} -4}{ \sqrt{2} } = \\ \\ \frac{ \sqrt{36}-4 }{ \sqrt{2} } = \\ \\ \frac{6-4}{ \sqrt{2} } = \\ \\ \frac{2}{ \sqrt{2} } = \\ \\ \frac{2. \sqrt{2} }{ \sqrt{2}. \sqrt{2} } = \frac{2. \sqrt{2} }{ \sqrt{2^2} } = \frac{\not2 \sqrt{2} }{\not2} =\fbox{ \sqrt{2} }$


c)
$\sqrt{2} - \frac{ \sqrt{2} }{1- \sqrt{2} } = \\ \\ \frac{ \sqrt{2}(1- \sqrt{2} )- \sqrt{2} }{1- \sqrt{2} } = \\ \\ \frac{ \sqrt{2}- \sqrt{2^2} - \sqrt{2} }{1- \sqrt{2} } = \\ \\ \frac{-2}{1- \sqrt{2} } = \\ \\ \frac{-2(1+ \sqrt{2}) }{(1_-\sqrt{2})(1+ \sqrt{2} } = \\ \\ \frac{-2(1+ \sqrt{2} )}{1^2- \sqrt{2^2} } = \\ \\ \frac{-2(1+ \sqrt{2}) }{1-2} = \\ \\ \frac{-2(1+ \sqrt{2} )}{-1} =2(1+ \sqrt{2} =\fbox{ 2+2 \sqrt{2} }$