Question

Rationalise 1/root3_root2 +1

Answer 1

Hey mate,..

\frac{1}{ \sqrt{3} - \sqrt{2} - 1}

= \frac{1}{( \sqrt{3} - \sqrt{2} ) - 1} \times \frac{ ( \sqrt{3} - \sqrt{2} ) + 1 }{ (\sqrt{3} - \sqrt{2} ) + 1}

= \frac{ (\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 }{3 + 2 - 2 \sqrt{6} - 1}

= \frac{ \sqrt{3} - \sqrt{2} + 1}{4 - 2 \sqrt{6} }

Now, rationalise this denominator..

= \frac{ \sqrt{3} - \sqrt{2} + 1}{4 - 2 \sqrt{6} } \times \frac{4 + 2 \sqrt{6} }{4 + 2 \sqrt{6} }

= \frac{( \sqrt{3} - \sqrt{2} + 1)(4 + 2 \sqrt{6}) }{(4 - 2 \sqrt{6})(4 + 2 \sqrt{6} ) }

= \frac{ 4 \sqrt{3} - 4 \sqrt{2} + 4 + 2 \sqrt{18} - 2 \sqrt{12} + 2 \sqrt{6} }{16 -24 }

\frac{4 \sqrt{3} - 4 \sqrt{2} + 4 + 18 \sqrt{2} - 4 \sqrt{3} + 2 \sqrt{6} }{ - 8}

= \frac{14 \sqrt{2} + 2 \sqrt{6} + 4 }{ - 8}

Hope it will help you.

✨Sai

Answer 2

heya !!! here is your answer .hope this helps you