Question

18. Rationalize the denominator 1 upon √3−√2+1​

Answer 1

Here is your answer :

\frac{1}{ \sqrt{3} - \sqrt{2} - 1}3​−2​−11​ 


= \frac{1}{( \sqrt{3} - \sqrt{2} ) - 1} \times \frac{ ( \sqrt{3} - \sqrt{2} ) + 1 }{ (\sqrt{3} - \sqrt{2} ) + 1}=(3​−2​)−11​×(3​−2​)+1(3​−2​)+1​ 


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


= \frac{ (\sqrt{3} - \sqrt{2} ) + 1}{ {( \sqrt{3} - \sqrt{2} ) }^{2} - {(1)}^{2} }=(3​−2​)2−(1)2(3​−2​)+1​ 


= \frac{ \sqrt{3} - \sqrt{2} + 1 }{3 + 2 - 2 \sqrt{6} - 1}=3+2−26​−13​−2​+1​ 


= \frac{ \sqrt{3} - \sqrt{2} + 1}{4 - 2 \sqrt{6} }=4−26​3​−2​+1​ 

Now, rationalise this denominator.. 

= \frac{ \sqrt{3} - \sqrt{2} + 1}{4 - 2 \sqrt{6} } \times \frac{4 + 2 \sqrt{6} }{4 + 2 \sqrt{6} }=4−26​3​−2​+1​×4+26​4+26​​ 


= \frac{( \sqrt{3} - \sqrt{2} + 1)(4 + 2 \sqrt{6}) }{(4 - 2 \sqrt{6})(4 + 2 \sqrt{6} ) }=(4−26​)(4+26​)(3​−2​+1)(4+26​)​ 


= \frac{ 4 \sqrt{3} - 4 \sqrt{2} + 4 + 2 \sqrt{18} - 2 \sqrt{12} + 2 \sqrt{6} }{16 -24 }=16−2443​−42​+4+218​−212​+26​​ 


\frac{4 \sqrt{3} - 4 \sqrt{2} + 4 + 18 \sqrt{2} - 4 \sqrt{3} + 2 \sqrt{6} }{ - 8}−843​−42​+4+182​−43​+26​​ 


= \frac{14 \sqrt{2} + 2 \sqrt{6} + 4 }{ - 8}=−8142​+26​+4​ 

Answer 2

1 / √3−√2+1

1/√3−√2+1 × 1/√3+√2-1/1/√3+√2-1

√3+√2-1 / 3+2+1

√3+√2-1/6

here's your answer

√3+√2-1/6

hope it helps mark me as brainliest