Question

give,√2=1.414 and √6=2.449,find the value of 1/√3-√2-1 correct to 3 palaces of decimals.​

Answer 1

$\frac{1}{ \sqrt{3} - \sqrt{2} - 1 } \\ \\ \\ \\ \frac{1( \sqrt{3} - \sqrt{2} + 1)}{( \sqrt{3} - \sqrt{2} - 1)( \sqrt{3} - \sqrt{2} + 1)} \\ \\ \\ \frac{ \sqrt{3} - \sqrt{2} + 1 }{ {( \sqrt{3} - \sqrt{2} )^{2} - {(1)}^{2} } } \\ \\ \\ \frac{ \sqrt{3} - \sqrt{2} + 1}{(5 - 2 \sqrt{6} ) - 1} \\ \\ \\ \frac{ \sqrt{3} - \sqrt{2} + 1}{4 - 2 \sqrt{6} } \\ \\ \\ \frac{ (\sqrt{3} - \sqrt{2} + 1)(4 + 2 \sqrt{6} )}{16 - 24} \\ \\ \\ \frac{4 \sqrt{3} - 4 \sqrt{2} + 4 + 2 \sqrt{18} - 2 \sqrt{12} + 2 \sqrt{6} }{ - 8} \\ \\ \\ \frac{2(2 \sqrt{3} - 2 \sqrt{2} + \sqrt{18} - \sqrt{12} + \sqrt{6} + 2 )}{ - 8} \\ \\ \\ \frac{ - 1(2 \sqrt{3} - 2 \sqrt{2} + \sqrt{18} - \sqrt{12} + \sqrt{6} + 2)}{4}$