Question

Ajutați mă vă rog ofer coroană și 50 de puncte ​

Answer 1

Explicație pas cu pas:

a)

$\frac{1}{ \sqrt{2} + 1} + \frac{1}{ \sqrt{3} + \sqrt{2} } + \frac{1}{ \sqrt{4} + \sqrt{3} } + \frac{1}{ \sqrt{5} + \sqrt{4} } + \frac{1}{ \sqrt{6} + \sqrt{5} } =$

$= \frac{\sqrt{2} - 1}{ (\sqrt{2} + 1)(\sqrt{2} - 1)} + \frac{\sqrt{3} - \sqrt{2}}{(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})} + \frac{\sqrt{4} - \sqrt{3}}{ (\sqrt{4} + \sqrt{3})(\sqrt{4} - \sqrt{3}) } + \frac{\sqrt{5} - \sqrt{4}}{ (\sqrt{5} + \sqrt{4})(\sqrt{5} - \sqrt{4}) } + \frac{\sqrt{6} - \sqrt{5}}{ (\sqrt{6} + \sqrt{5})(\sqrt{6} - \sqrt{5}) }$

$= \frac{\sqrt{2} - 1}{2 - 1} + \frac{\sqrt{3} - \sqrt{2}}{3 - 2} + \frac{\sqrt{4} - \sqrt{3}}{4 - 3} + \frac{\sqrt{5} - \sqrt{4}}{5 - 4} + \frac{\sqrt{6} - \sqrt{5}}{6 - 5}$

$= \sqrt{2} - 1 + \sqrt{3} - \sqrt{2} + \sqrt{4} - \sqrt{3} + \sqrt{5} - \sqrt{4} + \sqrt{6} - \sqrt{5}$

$= \sqrt{6} - 1$

b)

$\left( \frac{ \sqrt{3} }{2} - 3 \sqrt{2} + \frac{1}{ \sqrt{3} } + \frac{1}{ \sqrt{2} } \right) \div \left( \frac{1}{2 \sqrt{3} } - \frac{1}{ \sqrt{2} } \right) =$

$= \left( \frac{ \sqrt{3}\cdot  \sqrt{3} }{2 \sqrt{3} } - \frac{3 \sqrt{2}\cdot 2 \sqrt{3} }{2 \sqrt{3} } + \frac{2}{ \sqrt{3}\cdot 2} + \frac{\sqrt{6} }{ \sqrt{2}\cdot \sqrt{6} } \right) \div \left( \frac{1}{2 \sqrt{3}} - \frac{\sqrt{6} }{ \sqrt{2}\cdot \sqrt{6} } \right)$

$= \left( \frac{3 - 6 \sqrt{6} + 2 + \sqrt{6}}{2 \sqrt{3} } \right) \times \frac{2 \sqrt{3} }{1 - \sqrt{6} }$

$= \frac{5(1 - \sqrt{6}) }{2 \sqrt{3} } \times \frac{2 \sqrt{3} }{1 - \sqrt{6} } = 5$

c)

${(2 \sqrt{6} - 5)}^{2} + \frac{18}{ \sqrt{6} } - \frac{5}{ \sqrt{6} - 1} =$

$= 24 - 20 \sqrt{6} + 25 + \frac{18 \sqrt{6} }{6} - \frac{5( \sqrt{6} + 1)}{( \sqrt{6} - 1)( \sqrt{6} + 1)}$

$= 49 - 20 \sqrt{6} + 3 \sqrt{6} - \frac{5( \sqrt{6} + 1)}{6 - 1}$

$= 49 - 17 \sqrt{6} - ( \sqrt{6} + 1)$

$= 49 - 17 \sqrt{6} - \sqrt{6} - 1$

$= 48 - 18 \sqrt{6} = 6(8 - 3 \sqrt{6})$

e)

$\frac{1}{ \sqrt{3 + 2 \sqrt{2} } } + \frac{1}{ \sqrt{5 + 2 \sqrt{6} } } + \frac{1}{ \sqrt{7 + 2 \sqrt{12} } } + ... + \frac{1}{ \sqrt{4021 + 2 \sqrt{2010\cdot 2011} } }$

$= \frac{1}{ \sqrt{ {( \sqrt{1} )}^{2} + 2 \sqrt{1\cdot 2} + {( \sqrt{2} )}^{2} } } + \frac{1}{ \sqrt{ {( \sqrt{2} )}^{2} + 2 \sqrt{2\cdot 3} + {( \sqrt{3} )}^{2} }} + \frac{1}{ \sqrt{ {( \sqrt{3} )}^{2} + 2 \sqrt{3\cdot 4} + {( \sqrt{4} )}^{2} } } + ... + \frac{1}{ \sqrt{ {( \sqrt{2020} )}^{2} + 2 \sqrt{2010\cdot 2011} + {( \sqrt{2011} )}^{2} } }$

$= \frac{1}{ \sqrt{ {( \sqrt{1} + \sqrt{2})}^{2} } } + \frac{1}{ \sqrt{ {( \sqrt{2} + \sqrt{3})}^{2}}} + \frac{1}{ \sqrt{ {(\sqrt{3} + \sqrt{4} )}^{2} } } + ... + \frac{1}{ \sqrt{ {( \sqrt{2020} + \sqrt{2011} )}^{2} } }$

$= \frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + ... + \frac{1}{\sqrt{2020} + \sqrt{2011}}$

$= \frac{ \sqrt{2} - \sqrt{1} }{(\sqrt{2} + \sqrt{1})(\sqrt{2} - \sqrt{1})} + \frac{ \sqrt{3} - \sqrt{2} }{(\sqrt{3} - \sqrt{2})( \sqrt{3} + \sqrt{2} ) } + \frac{ \sqrt{4} - \sqrt{3} }{( \sqrt{4} + \sqrt{3})(\sqrt{4} - \sqrt{3})} + ... + \frac{ \sqrt{2011} - \sqrt{2010} }{(\sqrt{2021} + \sqrt{2010})( \sqrt{2011} - \sqrt{2010} )}$

$= \frac{ \sqrt{2} - \sqrt{1} }{1} + \frac{ \sqrt{3} - \sqrt{2} }{1} + \frac{ \sqrt{4} - \sqrt{3} }{1} + ... + \frac{ \sqrt{2011} - \sqrt{2010} }{1}$

$= \sqrt{2} - 1 + \sqrt{3} - \sqrt{2} + \sqrt{4} - \sqrt{3} + ... + \sqrt{2011} - \sqrt{2010}$

$= \sqrt{2011} - 1$