Math, asked by komalgoswami867, 4 months ago

simplify (minus 3 by root 3 + root 2) minus (3 root 2 by root 6 + root 3) + (4 root 3 by root 6 + root 2)



pls correct answers only ​

Answers

Answered by mathdude500
2

\large\underline{\bold{Given\:Question - }}

 \tt \: Simplify \:  :    \: \dfrac{ - 3}{ \sqrt{3}  +  \sqrt{2} }  - \dfrac{3 \sqrt{2} }{ \sqrt{6}  +  \sqrt{3} }  + \dfrac{4 \sqrt{3} }{ \sqrt{6 } +  \sqrt{2}  }

\large\underline{\bold{Solution :-  }}

Concept Used :-

Rationalization is the process of eliminating a radical or imaginary number from the denominator or numerator of an algebraic fraction. That is, remove the radicals in a fraction so that the denominator or numerator only contains a rational number.

The denominator of the above fraction has a binomial radical i.e., is the sum of two terms, one of which is an irrational number.

Multiply the numerator and denominator of the fraction with the conjugate of the radical.

Lets do it now!!

Now,

Consider

 \tt \: \rm :\longmapsto\:\dfrac{ 3}{ \sqrt{3}  +  \sqrt{2} }

On rationalizing the denominator, we get

\rm :\longmapsto\:\dfrac{ 3}{ \sqrt{3}  +  \sqrt{2} } \times \dfrac{ \sqrt{3}  -  \sqrt{2} }{ \sqrt{3}  -  \sqrt{2} }

\rm :\longmapsto\:\dfrac{3 \sqrt{3}  - 3 \sqrt{2} }{ {( \sqrt{3} )}^{2}  -  {( \sqrt{2}) }^{2} }

\rm :\longmapsto\:\dfrac{3 \sqrt{3}  - 3 \sqrt{2} }{1}

\rm :\implies\: \boxed{ \bf \: \dfrac{ 3}{ \sqrt{3}  +  \sqrt{2}} = 3 \sqrt{3} -  3 \sqrt{2}}

Now,

Consider,

\rm :\longmapsto\:\dfrac{4 \sqrt{3} }{ \sqrt{6 } +  \sqrt{2}  }

On rationalizing the denominator, we get

\rm :\longmapsto\:\dfrac{4 \sqrt{3} }{ \sqrt{6 } +  \sqrt{2}  }  \times \dfrac{ \sqrt{6}  -  \sqrt{2} }{ \sqrt{6} -  \sqrt{2}  }

\rm :\longmapsto\:\dfrac{4( \sqrt{18} -  \sqrt{6})  }{ {( \sqrt{6} )}^{2} -  {( \sqrt{2}) }^{2}  }

\rm :\longmapsto\:\dfrac{4(3 \sqrt{2}  -  \sqrt{6}) }{6 - 2}

\rm :\longmapsto\:3 \sqrt{2}  -  \sqrt{6}

\rm :\implies\: \boxed{ \bf \: \dfrac{4 \sqrt{3} }{ \sqrt{6 } +  \sqrt{2}  } = 3 \sqrt{2} -  \sqrt{6}   }

Now,

Consider,

\rm :\longmapsto\:\dfrac{3 \sqrt{2} }{ \sqrt{6}  +  \sqrt{3} }

On rationalizing the denominator, we get

\rm :\longmapsto\:\dfrac{3 \sqrt{2} }{ \sqrt{6}  +  \sqrt{3}}   \times \dfrac{ \sqrt{6} -  \sqrt{3}  }{ \sqrt{6} -  \sqrt{3}  }

\rm :\longmapsto\:\dfrac{3( \sqrt{12} -  \sqrt{6})  }{ {( \sqrt{6}) }^{2} -  {( \sqrt{3}) }^{2}  }

\rm :\longmapsto\:\dfrac{3(2 \sqrt{3} -  \sqrt{6} ) }{6 - 3}

\rm :\longmapsto\:\dfrac{3(2 \sqrt{3}  -  \sqrt{6}) }{3}

\rm :\implies\: \boxed{ \bf \: \dfrac{3 \sqrt{2} }{ \sqrt{6}  +  \sqrt{3} }   = 2 \sqrt{3}  -  \sqrt{6} }

So ,

\rm :\longmapsto\:\dfrac{ - 3}{ \sqrt{3}  +  \sqrt{2} } - \dfrac{3 \sqrt{2} }{ \sqrt{6}  +  \sqrt{3} }   + \dfrac{4 \sqrt{3} }{ \sqrt{6 } +  \sqrt{2}  }

\rm :\longmapsto\: - (3 \sqrt{3}  -  3 \sqrt{2} ) - (2 \sqrt{3}  -  \sqrt{6)}  + (3 \sqrt{2}  -  \sqrt{6)}

\rm :\longmapsto\: - 3 \sqrt{3}  +  3 \sqrt{2}  - 2 \sqrt{3}  +  \sqrt{6}  + 3 \sqrt{2}  -  \sqrt{6}

\rm :\implies\: -  5\sqrt{3}  + 6 \sqrt{2}

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