Question

find please i request u

Answer 1

${\bf HELLO \: \: DEAR,}$

$\bf \frac{2 + \sqrt{3}}{2 - \sqrt{3}} = a - b\sqrt{3} \\ \\ \bf now \: \: L.H.S, \\ \\ \bf \frac{2 + \sqrt{3}}{2 - \sqrt{3}} * \frac{2 + \sqrt{3}}{2 + \sqrt{3}} \\ \\ \bf \frac{(2 + \sqrt{3})^{2}}{{2}^{2} - {\sqrt{3}}^{2}} \\ \\ \bf \frac{4 + 3 + 4\sqrt{3}}{4 - 3} \\ \\ \bf \frac{7 + 4\sqrt{3}}{1} \\ \\ \bf on \: \: comparing \: \: with \: \: R.H.S, \\ \\ \bf [7 - (-4\sqrt{3})] = a - b\sqrt{3} \\ \\ \bf a = 7 , b = -4$

$\underline{\bf I \: \: HOPE \: \: ITS \: \: HELP \: \: YOU \: \: DEAR, \: \: THANKS}$

Answer 2

hiii!!!

here's Ur answer...

$given \: - > \frac{2 + \sqrt{3} }{2 - \sqrt{3} } = a - b \sqrt{3 } \\ \\ lhs = > \\ \\ \frac{2 + \sqrt{3} }{2 - \sqrt{3} } = \frac{2 + \sqrt{3} }{2 - \sqrt{3} } \times \frac{2 + \sqrt{3} }{2 + \sqrt{3} } \\ \\ = \frac{(2 + \sqrt{3} )(2 + \sqrt{3}) }{(2 - \sqrt{3})( 2 + \sqrt{3}) } \\ \\ = \frac{ {(2 + \sqrt{3}) }^{2} }{ {2}^{2} - {( \sqrt{3} )}^{2} } \\ \\ = \frac{ {(2)}^{2} + 2(2 \times \sqrt{3}) + {( \sqrt{3} )}^{2} }{4 - 3} \\ \\ = \frac{4 + 4 \sqrt{3} + 3}{1} \\ \\ = 7 + 4 \sqrt{3} \\ \\ rhs = > a - b \sqrt{3} \\ \\ therefore \: 7 + 4 \sqrt{3} = a - b \sqrt{3 } \\ \\ = > 7 - ( - 4 \sqrt{3} ) = a - b \sqrt{3} \\ \\ hence \: a \: = \: 7 \: and \: b \: = \: - 4 \sqrt{3} \\ \\ verification... \\ \\ lhs = > 7 + 4 \sqrt{3} \\ \\ rhs = > a - b \sqrt{3} \\ \\ = 7 - ( - 4 \sqrt{3} ) \\ \\ = 7 + 4 \sqrt{3} \\ \\ hence \: verified. \\ \\ lhs = rhs$

hope this helps..!!