Question

Algúem me ajuda por favor???

Racionalize o denominador de cada uma das seguintes expressões;
expressões em anexo

Answer 1

$a)\ \frac{2}{ \sqrt{7}+ \sqrt{5} } =\frac{2}{ \sqrt{7}+ \sqrt{5} }* \frac{ \sqrt{7}- \sqrt{5} }{ \sqrt{7}- \sqrt{5} } = \frac{2( \sqrt{7} - \sqrt{5} )}{( \sqrt{7})^2- (\sqrt{5})^2 }=\frac{2( \sqrt{7} - \sqrt{5} )}{ 7- 5 }=\frac{\diagup\!\!\!\!2( \sqrt{7} - \sqrt{5} )}{ \diagup\!\!\!\! 2 }=\\ \boxed{\boxed{\sqrt{7} - \sqrt{5}} }$


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$b)\ \frac{7}{ 3 - \sqrt{2} } =\frac{7}{ 3 - \sqrt{2} } * \frac{3+ \sqrt{2} }{3+ \sqrt{2} } = \frac{7(3+ \sqrt{2} )}{(3)^2- (\sqrt{2})^2 } =\frac{7(3+ \sqrt{2} )}{9-2} =\frac{\diagup\!\!\!\!7(3+ \sqrt{2} )}{\diagup\!\!\!\!7} =\boxed{\boxed{3+ \sqrt{2}} }$



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$c)\ \frac{1}{ \sqrt{7}-2 } =\frac{1}{ \sqrt{7}-2 }* \frac{\sqrt{7}+2}{\sqrt{7}+2} =\frac{\sqrt{7}+2}{ (\sqrt{7})^2-(2)^2 } =\frac{\sqrt{7}+2}{ 7-4 } =\boxed{\boxed{\frac{\sqrt{7}+2}{ 3 } }}$


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$d)\ \frac{ \sqrt{5} - \sqrt{2} }{ \sqrt{5}+ \sqrt{2} } = \frac{ \sqrt{5} - \sqrt{2} }{ \sqrt{5}+ \sqrt{2} } * \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} - \sqrt{2}} = \frac{ \sqrt{5} * \sqrt{5}- \sqrt{5 } * \sqrt{2}- \sqrt{2}* \sqrt{5} + \sqrt{2} * \sqrt{2} }{ (\sqrt{5})^2- (\sqrt{2} )^2 } = \\ \\ \frac{ \sqrt{25}- \sqrt{10}- \sqrt{10} + \sqrt{4} }{5-2} = \frac{5+2-2 \sqrt{10} }{3} = \boxed{\boxed{\frac{7-2 \sqrt{10}}{3} }}$


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$e)\ \frac{5+ \sqrt{11} }{3- \sqrt{11} } = \frac{5+ \sqrt{11} }{3- \sqrt{11} }* \frac{3+\sqrt{11}}{3+ \sqrt{11}} = \frac{15+5 \sqrt{11}+3 \sqrt{11}+ \sqrt{121} }{3^2-( \sqrt{11} )^2} = \frac{15+11+8 \sqrt{11} }{9-11} = \\ \\ \frac{26+8 \sqrt{11} }{-2} = \frac{\diagup\!\!\!\!2(13+4 \sqrt{11} )}{\diagup\!\!\!\!\!\!\!\!-2} =-(13+4 \sqrt{11} )=\boxed{\boxed{-13-4 \sqrt{11} }}$