Question

Rationalize the denominater of a) 2-√3 /√3
b) 1/5+√2
c) 5+√6/5-√6
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Answer 1

Answer:

$\frac{2 - \sqrt{3} }{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } = \frac{2 \sqrt{3} - 3 }{3}$

$\frac{1}{5 + \sqrt{2} } \times \frac{5 - \sqrt{2} }{5 - \sqrt{2} } = \frac{5 - \sqrt{2} }{25 - 2} = \frac{5 - \sqrt{2} }{23}$

$\frac{5 + \sqrt{6} }{5 - \sqrt{6} } \times \frac{5 + \sqrt{6} }{5 + \sqrt{6} } = \frac{25 + 6}{25 - 6} = \frac{31}{19}$

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Answer 2

$(1) \: \frac{2 - \sqrt{3} }{ \sqrt{3} } \\ = \frac{2 - \sqrt{3} }{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } \\ = \frac{(2 - \sqrt{3) } \sqrt{3} }{ { \sqrt{3} }^{2} } \\ = \frac{2 \sqrt{3 - 3} }{3}$

$(3) \: \frac{5 + \sqrt{6} }{5 - \sqrt{6} } \\ = \frac{5 + \sqrt{6} }{5 - \sqrt{6} } \times \frac{5 + \sqrt{6} }{5 + \sqrt{6} } \\ = \frac{(5 + \sqrt{6) ^{2} } }{ {5}^{2} - { \sqrt{6} }^{2} } \\ = \frac{ {5}^{2} + { \sqrt{6} }^{2} + 2 \times 5 \times \sqrt{6} }{25 - 6} \\ = \frac{25 + 6 + 10 \sqrt{6} }{19} \\ = \frac{31 + 10 \sqrt{6} }{19}$

$(2) \frac{1}{5 \sqrt{2} } \\ = \frac{1}{5 + \sqrt{2} } \times \frac{5 - \sqrt{2} }{5 - \sqrt{2} } \\ = \frac{5 - \sqrt{2} }{ {5}^{2} - { \sqrt{2} }^{2} } \\ = \frac{5 - \sqrt{2} }{25 - 2} \\ = \frac{5 - \sqrt{2} }{23}$