Question

answer in photo will be more appreciated​

Answer 1

Answer:

$\frac{1}{ \sqrt{2} + 1 } + \frac{1}{ \sqrt{3} + \sqrt{2} } + \frac{1}{2 + \sqrt{3} } = 1 \\ \frac{1}{ \sqrt{2} + 1 } \times \frac{ \sqrt{2} - 1 }{ \sqrt{2} - 1} + \frac{1}{ \sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} } + \frac{1}{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} } = 1 \\ \frac{1( \sqrt{2} - 1)}{( \sqrt{2} + 1)( \sqrt{2} - 1) } + \frac{1( \sqrt{3} - \sqrt{2} ) }{( \sqrt{3} + \sqrt{2})( \sqrt{3} - \sqrt{2})} + \frac{1(2 - \sqrt{3}) }{( 2 + \sqrt{3})(2 - \sqrt{3} ) } = 1 \\ \frac{ \sqrt{2} - 1 }{ {( \sqrt{2}) }^{2} - {(1)}^{2} } + \frac{ \sqrt{3} - \sqrt{2} }{ {( \sqrt{3}) }^{2} - {( \sqrt{2}) }^{2} } + \frac{2 - \sqrt{3} }{ {(2)}^{2} - {( \sqrt{3} )}^{2} } = 1 \\ \frac{ \sqrt{2} - 1}{2 - 1} + \frac{ \sqrt{3} - \sqrt{2} }{3 - 2} + \frac{2 - \sqrt{3} }{4 - 3} = 1 \\ \frac{ \sqrt{2} - 1}{1} + \frac{ \sqrt{3} - \sqrt{2} }{1} + \frac{2 - \sqrt{3} }{1} = 1 \\ \sqrt{2} - 1 + \sqrt{3} - \sqrt{2} + 2 - \sqrt{3} = 1 \\ \sqrt{2} - \sqrt{2} + \sqrt{3} - \sqrt{3} - 1 + 2 = 1 \\ 1 = 1 \\ LHS=RHS \\ Hence \: proved$

Answer 2

Answer:

4.page and last page of math question paper