Question

Rationalize 1/√2+√3-1

Answer 1

Given

$\dfrac{1}{\sqrt{2} } + \sqrt{3} -1$

Solving, we get,

$= \dfrac{1+\sqrt{6}-\sqrt{2} }{\sqrt{2} }$

$= \dfrac{1+\sqrt{6}-\sqrt{2} }{\sqrt{2} } * \dfrac{\sqrt{2} }{\sqrt{2} }$

To rationalise the denominator, multiply the numerator and  denominator with the root number.

$= \dfrac{\sqrt{2}(1+\sqrt{6}-\sqrt{2} ) }{2}$

$=\boxed{\boxed{\bold{ \dfrac{\sqrt{2}+2\sqrt{3}-2 }{2}}}}}$

                                                             

Answer 2

Question:

$Rationalize \: \frac{1}{ \sqrt{2} + \sqrt{3} - 1}$

Solution:

$= > \frac{1}{( \sqrt{2} + \sqrt{3} ) - 1} \times \frac{( \sqrt{2} + \sqrt{3}) + 1 }{( \sqrt{2} + \sqrt{3}) + 1} \\ \\ = > \frac{ \sqrt{2} + \sqrt{3} + 1 }{( \sqrt{2} + \sqrt{3} ) {}^{2} - (1) {}^{2} } \\ \\ using \: \: ( {x}^{2} - {y}^{2} ) = (x + y)(x - y) \\ \\ = > \frac{ \sqrt{2} + \sqrt{3} + 1 }{( \sqrt{2} ) {}^{2} + ( \sqrt{3} ) {}^{2} + 2( \sqrt{2} )( \sqrt{3} ) - 1 } \\ \\ = > \frac{ \sqrt{2} + \sqrt{3} + 1}{2 + 3 + 2 \sqrt{6} - 1 } \\ \\ = > \frac{ \sqrt{2} + \sqrt{3} + 1 }{5 - 1 + 2 \sqrt{6} } \\ \\ = > \frac{ \sqrt{2} + \sqrt{3} + 1 }{4 + 2 \sqrt{6} } \\ \\ = > \frac{ \sqrt{2} + \sqrt{3} + 1}{4 + 2 \sqrt{6} } \times \frac{4 - 2 \sqrt{6} }{4 - 2 \sqrt{6} } \\ \\ = > \frac{4( \sqrt{2} + \sqrt{3} + 1) - 2 \sqrt{6} ( \sqrt{2} + \sqrt{3} + 1) }{(4) {}^{2} - (2 \sqrt{6} ) {}^{2} } \\ \\ = > \frac{4 \sqrt{2} + 4 \sqrt{3} + 4 - 2 \sqrt{12} - 2 \sqrt{18} - 2 \sqrt{6} }{16 - 24} \\ \\ = > \frac{4 \sqrt{2} + 4 \sqrt{3} + 4 - 2 \sqrt{4 \times 3} - 2 \sqrt{9 \times 2} - 2 \sqrt{6} }{ - 8} \\ \\ = > we \: know \: that \: \sqrt{4} = 2 \: and \: \sqrt{9} = 9 \\ \\ = > \frac{4 \sqrt{2} + 4 \sqrt{3} + 4 - 4 \sqrt{3} - 6 \sqrt{2} - 2 \sqrt{6} }{ - 8} \\ \\ = > \frac{4 \sqrt{2} - 6 \sqrt{2} + \cancel{4 \sqrt{3}} \: - \cancel{ 4 \sqrt{3} } - 2 \sqrt{6} + 4}{ - 8 } \\ \\ = > \frac{ - 2 \sqrt{2} - 2 \sqrt{6} + 4}{ - 8} \\ \\ = > \frac{ - 2( \sqrt{2} + \sqrt{6} - 2)}{ - 8} \\ \\ = > \frac{ \cancel - 2( \sqrt{2} + \sqrt{6} - 2) }{ \cancel - 8} \\ \\ = > \frac{ \sqrt{2} + \sqrt{6} - 2}{4}$