Question

rationaloze √6+√3/√6-√3where √2=1.44

Answer 1

Answer:

$\frac{ \sqrt{6} + \sqrt{3} }{ \sqrt{6} - \sqrt{3} } = \frac{ \sqrt{6} + \sqrt{3} }{ \sqrt{6} - \sqrt{3} } \times \frac{ \sqrt{6} + \sqrt{3} }{ \sqrt{6} + \sqrt{3} } \\ = \frac{( \sqrt{6} + \sqrt{3}) ^{2} }{ { \sqrt{6} }^{2} - { \sqrt{3} }^{2} } \\ = \frac{ { (\sqrt{6}) }^{2} + ( { \sqrt{3} })^{2} + 2 \times \sqrt{6} \times \sqrt{3}}{6 - 3} \\ = \frac{6 + 3 + 2 \sqrt{18} }{3} \\ = \frac{9 + 2 \sqrt{9 \times 2} }{3} \\ = \frac{9 + 2 \times 3 \sqrt{2} }{3} \\ = \frac{9 + 6 \sqrt{2} }{3} \\ = \frac{3(3 + 2 \sqrt{2)} }{3} \\ = 3 + 2 \sqrt{2} \\ = 3 + 2 \times 1.414 \\ = 3 + 2.828 \\ = 5.828 \: \: \: ( \because \: \sqrt{2} = 1.414)$