Question

Va rog frumos ajutați-mă, dau coronița!.!!​

Answer 1

Răspuns:

$1)b) \frac{11}{ \sqrt{11} } = \frac{11 \times \sqrt{11} }{ \sqrt{11} \times \sqrt{11} } = \frac{11 \sqrt{11} }{11} = \sqrt{11} \\ f) \frac{ - 4}{ \sqrt{20} } = \frac{ - 4 \times \sqrt{20} }{ \sqrt{20} \times \sqrt{20} } = \frac{ - 4 \sqrt{20} }{20} = \frac{ \sqrt{20} }{ - 5} = \frac{ \sqrt{4 \times 5} }{ - 5} = \frac{2 \sqrt{5} }{ - 5}$

$j) \frac{ - 3}{2 \sqrt{6} } = \frac{ - 3 \times \sqrt{6} }{2 \sqrt{6} \times \sqrt{6} } = \frac{ - 3 \sqrt{6} }{12} = \frac{ \sqrt{6} }{ - 4}$

$3)b) \frac{3}{ \sqrt{5} - \sqrt{2} } = \frac{3 \times ( \sqrt{5} + \sqrt{2}) }{( \sqrt{5} - \sqrt{2} )( \sqrt{5} + \sqrt{2}) } = \frac{3 \sqrt{5} + 3 \sqrt{2} }{( { \sqrt{5} })^{2} - ( { \sqrt{2} })^{2} } = \frac{3 \sqrt{5} + 3 \sqrt{2} }{5 - 2} = \frac{ 3( \sqrt{5} + \sqrt{2} ) }{3} = \sqrt{5} + \sqrt{2}$

$f) \frac{18}{5 - \sqrt{13} } = \frac{18 \times (5 + \sqrt{13}) }{(5 - \sqrt{13})(5 + \sqrt{13} )} = \frac{18 \times (5 + \sqrt{13} )}{ {5}^{2} - ( { \sqrt{13} })^{2} } = \frac{18 \times (5 + \sqrt{13} )}{25 - 13} = \frac{18 \times (5 + \sqrt{13}) }{12} = \frac{3 \times (5 + \sqrt{13} )}{2} = \frac{15 + 3 \sqrt{13} }{2}$

$j) \frac{4}{5 + \sqrt{11} } = \frac{4 \times (5 - \sqrt{11} )}{(5 + \sqrt{11} )(5 - \sqrt{11}) } = \frac{4 \times (5 - \sqrt{11} )}{ {5}^{2} -( \sqrt{11}) ^{2} } = \frac{4 \times (5 - \sqrt{11}) }{25 - 11} = \frac{4 \times (5 - \sqrt{11} )}{14} = \frac{2 \times (5 - \sqrt{11}) }{7} = \frac{10 - 2 \sqrt{11} }{7}$

$n) \frac{ - 2 \sqrt{15} }{ \sqrt{5} + \sqrt{3} } = \frac{ - 2 \sqrt{15} \times ( \sqrt{5} - \sqrt{3}) }{( \sqrt{5} + \sqrt{3})( \sqrt{5} - \sqrt{3}) } = \frac{ - 2 \sqrt{15} \times ( \sqrt{5} - \sqrt{3} )}{( { \sqrt{5} })^{2} - ( { \sqrt{3} })^{2} } = \frac{ - 2 \sqrt{15} \times ( \sqrt{5} - \sqrt{3} ) }{5 - 3} = \frac{ - 2 \sqrt{15} \times ( \sqrt{5} - \sqrt{3} ) }{2} = - \sqrt{15} \times ( \sqrt{5} - \sqrt{3} ) = \\ = - \sqrt{75} + \sqrt{45} = - \sqrt{25 \times 3} + \sqrt{9 \times 5} = - 5 \sqrt{3} + 3 \sqrt{5}$

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