Question

Urgentt, va rog frumos!! ​

Answer 1

Răspuns:

$a)a = \sqrt{3 + 2 \sqrt{2} } \\ b = \sqrt{3 - 2 \sqrt{2} } \\ a \times b = \sqrt{3 + 2 \sqrt{2} } \times \sqrt{3 - 2 \sqrt{2} } = \sqrt{ {3}^{2} - ( {2 \sqrt{2} })^{2} } = \sqrt{9 - 4 \times 2} = \sqrt{9 - 8} = \sqrt{1} = 1$

$b) { a}^{2} - 2ab + {b}^{2} = ( { \sqrt{3 + 2 \sqrt{2} } })^{2} - 2 \times \sqrt{3 + 2 \sqrt{2} } \times \sqrt{3 - 2 \sqrt{2} } + ( { \sqrt{3 - 2 \sqrt{2} } })^{2} = |3 + 2 \sqrt{2} | - 2 \times 1 + |3 - 2 \sqrt{2} | = |3 + \sqrt{ {2}^{2} \times 2 } | - 2 + |3 - \sqrt{ {2}^{2} \times 2 } | = \\ = |3 + \sqrt{8} | - 2 + |3 - \sqrt{8} | = | \sqrt{9} + \sqrt{8} | - 2 + | \sqrt{9} - \sqrt{8} | \\ cum \: \sqrt{9} > \sqrt{8} rezulta \: ca \: | \sqrt{9} + \sqrt{8} | > 0 \: rezulta \: ca \: | \sqrt{9} + \sqrt{8} | = \sqrt{9} + \sqrt{8} \\ \sqrt{9} > \sqrt{8} rezulta \: ca \: | \sqrt{9} - \sqrt{8} | > 0 \: rezulta \: ca \: | \sqrt{9} - \sqrt{8} | = \sqrt{9} - \sqrt{8} \\ {a}^{2} - 2ab + {b}^{2} = \sqrt{9} + \sqrt{8} - 2 + \sqrt{9} - \sqrt{8} = 3 - 2 + 3 = 4$

$c) \frac{1}{a} - \frac{1}{b} = \frac{1}{ \sqrt{3 + 2 \sqrt{2} } } - \frac{1}{ \sqrt{3 - 2 \sqrt{2} } } = \frac{ \sqrt{3 - 2 \sqrt{2} } }{ (\sqrt{3 + 2 \sqrt{2} } )( \sqrt{3 - 2 \sqrt{2} }) } - \frac{ \sqrt{3 + 2 \sqrt{2} } }{( \sqrt{3 - 2 \sqrt{2}) } ( \sqrt{3 + 2 \sqrt{2}) } } = \\ = \frac{ \sqrt{3 - 2 \sqrt{2} } }{1} - \frac{ \sqrt{3 + 2 \sqrt{2} } }{1} = \sqrt{3 - 2 \sqrt{2} } - \sqrt{3 + 2 \sqrt{2} } = \sqrt{3 - \sqrt{8} } - \sqrt{3 + \sqrt{8} }$