Question

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Answer 1

$a, (\frac{2}{\sqrt{3} - 1} + \frac{3}{\sqrt{3} - 2} + \frac{15}{3 - \sqrt{3}}) . \frac{1}{\sqrt{3} + 5} = [\frac{2 . (\sqrt{3} + 1)}{(\sqrt{3} - 1) (\sqrt{3} + 1)} + \frac{3 . (\sqrt{3} + 2)}{(\sqrt{3} - 2) (\sqrt{3} + 2)} + \frac{15 . (3 + \sqrt{3})}{(3 - \sqrt{3}) . (3 + \sqrt{3})}] . \frac{1}{\sqrt{3} + 5} = (\sqrt{3} + 1 - 3 . (\sqrt{3} + 2) + \frac{5}{2} . (3 + \sqrt{3})) . \frac{1}{\sqrt{3} + 5} = (\sqrt{3} + 1 - 3 \sqrt{3} - 6 + \frac{15}{2} + \frac{5}{2} \sqrt{3}) . \frac{1}{\sqrt{3} + 5} = (\frac{1}{2} \sqrt{3} + \frac{5}{2}) . \frac{1}{\sqrt{3} + 5} = \frac{1}{2} . (\sqrt{3} + 5) . \frac{1}{\sqrt{3} + 5} = \frac{1}{2}$

$b, (\frac{\sqrt{14} - \sqrt{7}}{1 - \sqrt{2}} + \frac{\sqrt{15} - \sqrt{5}}{1 - \sqrt{3}}) : \frac{1}{\sqrt{7} - \sqrt{5}} = [\frac{\sqrt{7} . (\sqrt{2} - 1)}{1 - \sqrt{2}} + \frac{\sqrt{5} . (\sqrt{3} - 1)}{1 - \sqrt{3}}] . (\sqrt{7} - \sqrt{5}) = (- \sqrt{7} - \sqrt{5}) . (\sqrt{7} - \sqrt{5}) = - (7 - 5) = - 2$ .

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Answer 2

$a, (\frac{2}{\sqrt{3} - 1} + \frac{3}{\sqrt{3} - 2} + \frac{15}{3 - \sqrt{3}}) . \frac{1}{\sqrt{3} + 5} = [\frac{2 . (\sqrt{3} + 1)}{(\sqrt{3} - 1) (\sqrt{3} + 1)} + \frac{3 . (\sqrt{3} + 2)}{(\sqrt{3} - 2) (\sqrt{3} + 2)} + \frac{15 . (3 + \sqrt{3})}{(3 - \sqrt{3}) . (3 + \sqrt{3})}] . \frac{1}{\sqrt{3} + 5} = (\sqrt{3} + 1 - 3 . (\sqrt{3} + 2) + \frac{5}{2} . (3 + \sqrt{3})) . \frac{1}{\sqrt{3} + 5} = (\sqrt{3} + 1 - 3 \sqrt{3} - 6 + \frac{15}{2} + \frac{5}{2} \sqrt{3}) . \frac{1}{\sqrt{3} + 5} = (\frac{1}{2} \sqrt{3} + \frac{5}{2}) . \frac{1}{\sqrt{3} + 5} = \frac{1}{2} . (\sqrt{3} + 5) . \frac{1}{\sqrt{3} + 5} = \frac{1}{2}$

$b, (\frac{\sqrt{14} - \sqrt{7}}{1 - \sqrt{2}} + \frac{\sqrt{15} - \sqrt{5}}{1 - \sqrt{3}}) : \frac{1}{\sqrt{7} - \sqrt{5}} = [\frac{\sqrt{7} . (\sqrt{2} - 1)}{1 - \sqrt{2}} + \frac{\sqrt{5} . (\sqrt{3} - 1)}{1 - \sqrt{3}}] . (\sqrt{7} - \sqrt{5}) = (- \sqrt{7} - \sqrt{5}) . (\sqrt{7} - \sqrt{5}) = - (7 - 5) = - 2$ .

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