Math, asked by fazilansari, 1 year ago

find please i request u

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Answered by rohitkumargupta
6
{\bf HELLO \: \: DEAR,}

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

\underline{\bf I \: \: HOPE \: \: ITS \: \: HELP \: \: YOU \: \: DEAR,<br />\: \: THANKS}

rohitkumargupta: :-)
fazilansari: yes dost cool
fazilansari: brother
Answered by Anonymous
7
hiii!!!

here's Ur answer...

given \:  -  &gt;  \frac{2 +  \sqrt{3} }{2 -  \sqrt{3} }  = a - b \sqrt{3 }  \\  \\  lhs =  &gt;  \\  \\  \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  =  &gt; a - b \sqrt{3}  \\  \\ therefore \: 7 + 4 \sqrt{3}  = a - b \sqrt{3 }  \\  \\  =  &gt; 7 - ( - 4 \sqrt{3} ) = a - b \sqrt{3}  \\  \\ hence \: a \:  =  \: 7 \: and \: b \:  =  \:  - 4 \sqrt{3}  \\  \\ verification... \\  \\ lhs  =  &gt; 7 + 4 \sqrt{3}  \\  \\ rhs =  &gt; a - b \sqrt{3}  \\  \\  = 7 - ( - 4 \sqrt{3} ) \\  \\ = 7 + 4 \sqrt{3}  \\  \\ hence \: verified. \\  \\ lhs = rhs

hope this helps..!!

Anonymous: thank u for brainliest :)
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