Question

can anyone help me <br />$express \: with \: rational \: denominator \\ \frac{1}{( \sqrt{2 \: - \sqrt{3) + \sqrt{4} } } }$<br />​

Answer 1

Answer:

$\frac{1}{ \sqrt{2} - \sqrt{3} + \sqrt{4} } \\ \\ \frac{1}{ (\sqrt{2} - \sqrt{3}) + 2 } \times \frac{( \sqrt{2} - \sqrt{3}) - 2 }{( \sqrt{2} - \sqrt{3}) - 2} \\ \\ \frac{ \sqrt{2} - \sqrt{3} - 2}{( \sqrt{2} - \sqrt{3})^{2} - (2)^{2} } \\ \\ \frac{ \sqrt{2} - \sqrt{3} - 2 }{2 + 3 - 2 \sqrt{6} - 4 } \\ \\ \frac{ \sqrt{2} - \sqrt{3} - 2 }{1 - 2 \sqrt{6} }$

Still the denominator is not rationalised, we will rationalise again :

$\frac{ \sqrt{2} - \sqrt{3} - 2 }{1 - 2 \sqrt{6} } \\ \\\frac{ \sqrt{2} - \sqrt{3} - 2}{1 - 2 \sqrt{6} } \times \frac{1 + 2 \sqrt{6} }{1 + 2 \sqrt{6} } \\ \\ \frac{(1 + 2 \sqrt{6})( \sqrt{2} - \sqrt{3} - 2)}{(1) {}^{2} - (2 \sqrt{6}) {}^{2} } \\ \\ \frac{ \sqrt{2} - \sqrt{3} - 2 + 4\sqrt{3} -6 \sqrt{2} - 4 \sqrt{6} }{1 - 24} \\ \\ \frac{-5 \sqrt{2} + 3 \sqrt{3} - 2- 4 \sqrt{6} }{-23}\\ \\ \frac{-(5 \sqrt{2} - 3 \sqrt{3} + 4 \sqrt{6}+2 }{-23} \\ \\ \frac{5 \sqrt{2}- 3 \sqrt{3} + 4 \sqrt{6} +2}{23}$