Math, asked by prince5132, 11 months ago

please solve both 1 and 2
please
please

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Answers

Answered by Mysterioushine
24

\huge\tt{\underline{\underline{Solution:-}}}

(i) \:  \:  \:  \frac{3 \sqrt{2} - 2 \sqrt{3}  }{3 \sqrt{2} + 2 \sqrt{3}  }  +  \frac{ \sqrt{12} }{ \sqrt{3}  -  \sqrt{2} }  \\  \\  = \:  \frac{(3 \sqrt{2}  - 2 \sqrt{3})( \sqrt{3}  -  \sqrt{2}) +  2\sqrt{3}  (3 \sqrt{2}   + 2 \sqrt{3}) }{(3 \sqrt{2} + 2 \sqrt{3})( \sqrt{3} -  \sqrt{2} )   }  \\  \\  =   \frac{(3 \sqrt{6} - 6 - 6 + 2\sqrt{6} ) + ( 6\sqrt{6}   + 12)}{(3 \sqrt{6} - 6 + 6 - 2 \sqrt{6} ) }  \\  \\  = \frac{(5 \sqrt{6} - 12 + 6 \sqrt{6}  + 12)  }{ \sqrt{6} }    \\  \\  = \:  \frac{11 \sqrt{6} }{ \sqrt{6} }   = 11

 (ii) \:  \frac{7 + 3 \sqrt{5} }{3 +  \sqrt{5} }  -  \frac{7 -  3\sqrt{5} }{3 -   \sqrt{5} } \\  \\  =  \frac{(7 + 3 \sqrt{5})(3 -  \sqrt{5}) - (7 -  3\sqrt{5})(3 +  \sqrt{5}  )  }{(3 +  \sqrt{5})(3 -  \sqrt{5} ) }   \\  \\  =  \frac{21 - 7 \sqrt{5}  + 9 \sqrt{5}  - 15 -(21 + 7 \sqrt{5} - 9 \sqrt{5}  - 15)  }{9 - 5}  \\  \\  =   \frac{21 - 7 \sqrt{5} + 9 \sqrt{5}   - 15 - 21 - 7 \sqrt{5}  + 9 \sqrt{5} + 15 }{4} \\  \\   =  \frac{ - 14 \sqrt{5} + 18 \sqrt{5}  }{4}  =  \frac{4 \sqrt{5} }{ 4 }  =  \sqrt{5}

Answered by Rohith200422
21

Question:

(i) \dfrac{3 \sqrt{2} - 2 \sqrt{3}   }{3 \sqrt{2} + 2 \sqrt{3}  }  +  \dfrac{ \sqrt{12} }{3 \sqrt{2}  - 2 \sqrt{3} }

(ii) \dfrac{7 + 3 \sqrt{5} }{3 +  \sqrt{5} }  -  \dfrac{7 - 3 \sqrt{5} }{3 -  \sqrt{5} }

Answer:

(i) \:   \bold{\:11\: }  }

(ii) \:   \bf{ \: \sqrt{5}\: }

Step-by-step explanation:

(i) \dfrac{3 \sqrt{2} - 2 \sqrt{3} }{3 \sqrt{2} + 2 \sqrt{3} } + \dfrac{ \sqrt{12} }{ \sqrt{3} - \sqrt{2} }

\implies \dfrac{(3 \sqrt{2} - 2 \sqrt{3})( \sqrt{3} - \sqrt{2}) + 2\sqrt{3} (3 \sqrt{2} + 2 \sqrt{3}) }{(3 \sqrt{2} + 2 \sqrt{3})( \sqrt{3} - \sqrt{2} ) }

\implies \dfrac{(3 \sqrt{6} - 6 - 6 + 2\sqrt{6} ) + ( 6\sqrt{6} + 12)}{(3 \sqrt{6} - 6 + 6 - 2 \sqrt{6} ) }

\implies \dfrac{(5 \sqrt{6} - 12 + 6 \sqrt{6} + 12) }{ \sqrt{6} }

\implies \dfrac{11 \sqrt{6} }{ \sqrt{6} }

\implies boxed{11 }

 \underline{ \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:\:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: }

(ii) \: \dfrac{7 + 3 \sqrt{5} }{3 + \sqrt{5} } - \frac{7 - 3\sqrt{5} }{3 - \sqrt{5} }

\longmapsto \dfrac{(7 + 3 \sqrt{5})(3 - \sqrt{5}) - (7 - 3\sqrt{5})(3 + \sqrt{5} ) }{(3 + \sqrt{5})(3 - \sqrt{5} ) }

\longmapsto \dfrac{21 - 7 \sqrt{5} + 9 \sqrt{5} - 15 -(21 + 7 \sqrt{5} - 9 \sqrt{5} - 15) }{9 - 5}

 \longmapsto\dfrac{21 - 7 \sqrt{5} + 9 \sqrt{5} - 15 - 21 - 7 \sqrt{5} + 9 \sqrt{5} + 15 }{4}

\longmapsto \dfrac{ - 14 \sqrt{5} + 18 \sqrt{5} }{4}

 \dfrac{4 \sqrt{5} }{ 4 }

\boxed{ \sqrt{5}}

 \underline{ \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: }

Identities used :

 \bigstar \:  {a}^{2}  -  {b}^{2}  = (a - b)(a + b)

Calculations :

 \star \: 3 \sqrt{2}  =  \sqrt{3}  \times  \sqrt{3}  \times  \sqrt{2}

\star \: \sqrt{12}  = 2 \sqrt{6}

\star \: \sqrt{12}  \times  \sqrt{6}  = 2 \sqrt{3}  \times  \sqrt{2}  \times  \sqrt{6}

\star \:  \sqrt{12}  \times  \sqrt{6}  = 6 \sqrt{2}

\star \: \sqrt{6}  \big( \sqrt{3}  -  \sqrt{2} \big ) = 3 \sqrt{2}  - 2 \sqrt{3}

\star \:2 \sqrt{3}  =  \sqrt{2}  \times  \sqrt{2}   \times  \sqrt{3}

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