Question

a soma $\frac{1}{ \sqrt{5} + \sqrt{8} } }+ \frac{1}{ \sqrt{8}+ \sqrt{11} } + \frac{1}{ \sqrt{11}+ \sqrt{14} } + \frac{1}{ \sqrt{14}+ \sqrt{17} } + \frac{1}{ \sqrt{17}+ \sqrt{20} }$ é igual a?

Answer 1

$\dfrac{1}{\sqrt{5}+\sqrt{8}}}+\dfrac{1}{\sqrt{8}+\sqrt{11}}+\dfrac{1}{\sqrt{11}+\sqrt{14}}+\dfrac{1}{ \sqrt{14}+\sqrt{17}}+\dfrac{1}{\sqrt{17}+ \sqrt{20}}$

$\dfrac{1}{\sqrt{5}+\sqrt{8}}*\dfrac{\sqrt{5}-\sqrt{8}}{\sqrt{5}-\sqrt{8}}$
$\dfrac{\sqrt{5}-\sqrt{8}}{5-8}$
$\dfrac{\sqrt{5}-\sqrt{8}}{-3}$

$\dfrac{1}{\sqrt{8}+\sqrt{11}}*\dfrac{\sqrt{8}-\sqrt{11}}{\sqrt{8}-\sqrt{11}}$
$\dfrac{\sqrt{8}-\sqrt{11}}{8-11}$
$\dfrac{\sqrt{8}-\sqrt{11}}{-3}$

$\dfrac{1}{\sqrt{11}+\sqrt{14}}*\dfrac{\sqrt{11}-\sqrt{14}}{\sqrt{11}-\sqrt{14}}$
$\dfrac{\sqrt{11}-\sqrt{14}}{11-14}$
$\dfrac{\sqrt{11}-\sqrt{14}}{-3}$

$\dfrac{1}{\sqrt{14}+\sqrt{17}}*\dfrac{\sqrt{14}-\sqrt{17}}{\sqrt{14}-\sqrt{17}}$
$\dfrac{\sqrt{14}-\sqrt{17}}{14-17}$
$\dfrac{\sqrt{14}-\sqrt{17}}{-3}$

$\dfrac{1}{\sqrt{17}+\sqrt{20}}*\dfrac{\sqrt{17}-\sqrt{20}}{\sqrt{17}-\sqrt{20}}$
$\dfrac{\sqrt{17}-\sqrt{20}}{17-20}$
$\dfrac{\sqrt{17}-\sqrt{20}}{-3}$


$\dfrac{(\sqrt{5}-\sqrt{8})-(\sqrt{8}-\sqrt{11})-(\sqrt{11}-\sqrt{14})-(\sqrt{14}-\sqrt{17})-\sqrt{17}-\sqrt{20}}{3}=$
$\dfrac{-\sqrt{5}+\sqrt{20}}{3}=$
$\dfrac{-\sqrt{5}+\sqrt{2^2*5}}{3}=$
$\dfrac{-\sqrt{5}+2\sqrt{5}}{3}=$
$\boxed{\boxed{\dfrac{\sqrt{5}}{3}}}$

Answer 2

Olá Yuri.


Organizando as frações e racionalizando uma a uma.


$\mathsf{\star~\boxed{\boxed{\mathsf{a^2+b^2=(a+b)\cdot(a-b)}}}}$


$\mathsf{\dfrac{1}{\sqrt{5}+\sqrt{8}}\cdot\dfrac{\sqrt{5}-\sqrt{8}}{\sqrt{5}-\sqrt{8}}\Rightarrow\dfrac{\sqrt{5}-\sqrt{8}}{5-8}\Rightarrow\dfrac{\sqrt{5}-\sqrt{8}}{-3}}$

$\mathsf{\dfrac{1}{\sqrt{8}+\sqrt{11}}\cdot\dfrac{\sqrt{8}-\sqrt{11}}{\sqrt{8}-\sqrt{11}}\Rightarrow\dfrac{\sqrt{8}-\sqrt{11}}{8-11}\Rightarrow\dfrac{\sqrt{8}-\sqrt{11}}{-3}}\\\\\\\mathsf{\dfrac{1}{\sqrt{11}+\sqrt{14}}\cdot\dfrac{\sqrt{11}-\sqrt{14}}{\sqrt{11}-\sqrt{14}}\Rightarrow\dfrac{\sqrt{11}-\sqrt{14}}{11-14}\Rightarrow\dfrac{\sqrt{11}-\sqrt{14}}{-3}}$

$\mathsf{\dfrac{1}{\sqrt{14}+ \sqrt{17}}\cdot\dfrac{\sqrt{14}-\sqrt{17}}{\sqrt{14}-\sqrt{17}}\Rightarrow\dfrac{\sqrt{14}-\sqrt{17}}{14-17}\Rightarrow\dfrac{\sqrt{14}-\sqrt{17}}{-3}}\\\\\\\mathsf{\dfrac{1}{\sqrt{17}+\sqrt{20}}\cdot\dfrac{\sqrt{17}-\sqrt{20}}{\sqrt{17}-\sqrt{20}}\Rightarrow\dfrac{\sqrt{17}-\sqrt{20}}{17-20}\Rightarrow\dfrac{\sqrt{17}-\sqrt{20}}{-3}}$

Organizando as frações.

$\mathsf{\dfrac{-(\sqrt{5}-\!\!\sqrt{8})-(\sqrt{8}-\!\!\sqrt{11})-(\sqrt{11}-\!\!\sqrt{14})-(\sqrt{14}-\!\!\sqrt{17})-\sqrt{17}-\!\!\sqrt{20}}{3}}\\\\\\\mathsf{\dfrac{\diagup\!\!\!\!\!\!\!\sqrt{8}-\sqrt{5}+\sqrt{\diagdown\!\!\!\!\!11}-\sqrt{\diagup\!\!\!\!8}+\sqrt{\diagdown\!\!\!\!14}-\sqrt{\diagdown\!\!\!\!\!11}+\sqrt{\diagup\!\!\!\!17}-\sqrt{\diagdown\!\!\!\!14}+\sqrt{20}-\sqrt{\diagup\!\!\!\!\!17}}{3}}\\\\\mathsf{\dfrac{-\sqrt{5}+\sqrt{20}}{3}}\\\\\\\mathsf{\dfrac{\sqrt{2^2\cdot5}-\sqrt{5}}{3}}$

$\mathsf{\dfrac{2\sqrt{5}-\sqrt{5}}{3}}\\\\\\\boxed{\mathsf{\dfrac{\sqrt{5}}{3}}}$


Dúvidas? comente.