tolong bantu kakk pake cara.... Question from @ClaraAyyana

Bagian A

$\boxed{\begin{aligned}\sf \frac{2}{2 + \sqrt{2} } &=\sf \frac{2}{2 + \sqrt{2} } \times \frac{ 2 - \sqrt{2} }{ 2 - \sqrt{2} } \\ \sf &=\sf \frac{2(2 - \sqrt{2}) }{4 - 2} \\ \sf &= \sf \frac{2(2 - \sqrt{2} )}{2} \\ \sf &=\sf 2 - \sqrt{2} \end{aligned}}$

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Bagian B

$\boxed{\begin{aligned}\sf \frac{3}{4 - \sqrt{3} } &=\sf \frac{3}{4 - \sqrt{3} } \times \frac{4 + \sqrt{3} }{4 + \sqrt{3} } \\ \sf &=\sf \frac{3(4 + \sqrt{3}) }{16 - 3} \\ \sf &= \sf \frac{12 + 3\sqrt{3} }{13}\end{aligned}}$

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Bagian C

$\boxed{\begin{aligned}\sf \frac{4}{3 - \sqrt{5} } &=\sf \frac{4}{3 - \sqrt{5} } \times \frac{3 + \sqrt{5} }{3 + \sqrt{5} } \\ \sf &=\sf \frac{4(3 + \sqrt{5} )}{9 - 5} \\ \sf &= \sf \frac{4(3 + \sqrt{5} )}{4} \\ \sf &=\sf 3 + \sqrt{5} \end{aligned}}$

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Bagian D

$\boxed{\begin{aligned}\sf \frac{10}{6 - \sqrt{6} } &=\sf \frac{10}{6 - \sqrt{6} } \times \frac{6 + \sqrt{6} }{6 + \sqrt{6} } \\ \sf &=\sf \frac{10(6 + \sqrt{6} )}{36 - 6} \\ \sf &= \sf \frac{10(6 + \sqrt{6} )}{30} \\ \sf &=\sf \frac{6 + \sqrt{6} }{3} \end{aligned}}$

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Bagian E

$\boxed{\begin{aligned}\sf \frac{2 \sqrt{3} }{3 - \sqrt{6} } &=\sf \frac{2 \sqrt{3} }{3 - \sqrt{6} } \times \frac{3 + \sqrt{6} }{3 + \sqrt{6} } \\ \sf &=\sf \frac{2 \sqrt{3} \: (3 + \sqrt{6} )}{9 - 6} \\ \sf &= \sf \frac{6 \sqrt{3} + 2 \sqrt{18} }{3} \\ \sf &=\sf \frac{6 \sqrt{3} + 2 \sqrt{9} \sqrt{2} }{3}\\ \sf &=\sf \frac{6 \sqrt{3} + 6 \sqrt{2} }{3}\\ \sf &=\sf \frac{3(2 \sqrt{3} + 2 \sqrt{2} ) }{3}\\ \sf &=\sf2 \sqrt{3} + 2 \sqrt{2} \end{aligned}}$

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Bagian F

$\boxed{\begin{aligned}\sf \frac{4 \sqrt{5} }{3 - \sqrt{5} } &=\sf \frac{4 \sqrt{5} }{3 - \sqrt{5} } \times \frac{3 + \sqrt{5} }{3 + \sqrt{5} } \\ \sf &=\sf \frac{4 \sqrt{5} \: (3 + \sqrt{5} )}{9 - 5} \\ \sf &= \sf \frac{4 \sqrt{5} \: (3 + \sqrt{5} )}{4}\\ \sf &=\sf \sqrt{5} \: (3 + \sqrt{5} )\\ \sf &=\sf3 \sqrt{5} + 5 \end{aligned}}$