Question

Dau coroana!!!!!!!!!!!!!!!!!!!!

Answer 1

Explicație pas cu pas:

4

$\dfrac{1}{ \sqrt{6} - 1} = \dfrac{\sqrt{6} + 1}{ (\sqrt{6} - 1)(\sqrt{6} + 1)} = \dfrac{\sqrt{6} + 1}{6 - 1} = \dfrac{\sqrt{6} + 1}{5}$

$\dfrac{2}{ \sqrt{6} - 2} = \dfrac{2(\sqrt{6} + 2)}{ (\sqrt{6} - 2)(\sqrt{6} + 2)} = \dfrac{2(\sqrt{6} + 2)}{6 - 4} = \dfrac{2(\sqrt{6} + 2)}{2} = \sqrt{6} + 2$

$\dfrac{ \sqrt{3} }{ \sqrt{5} - \sqrt{3} } = \dfrac{ \sqrt{3} (\sqrt{5} + \sqrt{3} )}{ (\sqrt{5} - \sqrt{3} )(\sqrt{5} + \sqrt{3} )} = \dfrac{ \sqrt{15} + 3}{5 - 3} = \dfrac{3 + \sqrt{15}}{2}$

$\dfrac{2}{ \sqrt{3} - \sqrt{2} } = \dfrac{2(\sqrt{3} + \sqrt{2} )}{ (\sqrt{3} - \sqrt{2} )(\sqrt{3} + \sqrt{2} )} = \dfrac{2(\sqrt{3} + \sqrt{2} )}{3 - 2} = \dfrac{2(\sqrt{3} + \sqrt{2} )}{1} = 2(\sqrt{3} + \sqrt{2}$

$\dfrac{1}{ \sqrt{5} + 1} = \dfrac{\sqrt{5} - 1}{ (\sqrt{5} + 1)(\sqrt{5} - 1)} = \dfrac{\sqrt{5} - 1}{5 - 1} = \dfrac{\sqrt{5} - 1}{4}$

5.

$\sqrt{( \sqrt{3} - \sqrt{2} )( \sqrt{3} + \sqrt{2} )} = \sqrt{ {( \sqrt{3} )}^{2} - {( \sqrt{2} )}^{2} } = \sqrt{3 - 2} = \sqrt{1} = 1$

$\sqrt{( \sqrt{10} + 1)( \sqrt{10} - 1)} = \sqrt{ {( \sqrt{10})}^{2} - {1}^{2} } = \sqrt{10 - 1} = \sqrt{9} = 3$

$\sqrt{(6 - \sqrt{5})(6 + \sqrt{5} )} = \sqrt{ {6}^{2} - {( \sqrt{5} )}^{2} } = \sqrt{36 - 5} = \sqrt{31}$

$\sqrt{(2 \sqrt{3} - 1)(2 \sqrt{3} + 1)} = \sqrt{ {(12)}^{2} - {1}^{2} } = \sqrt{12 - 1} = \sqrt{11}$

$\sqrt{(3 \sqrt{2} + 2 \sqrt{3})(3 \sqrt{2} - 2 \sqrt{3})} = \sqrt{ {( \sqrt{18} )}^{2} - {( \sqrt{12} )}^{2} } = \sqrt{18 - 12} = \sqrt{6}$