Math, asked by enonchloe3, 5 hours ago

rationalise the denominator​

Attachments:

Answers

Answered by SachinGupta01
57

\underline{\underline{\sf{\maltese\:\: Solution }}}

 \sf(a) \: \:  \:  \dfrac{1}{ \sqrt{3} -  \sqrt{2}  }

On rationalising,

 \sf \implies  \dfrac{1}{ \sqrt{3} -  \sqrt{2}  }   \times  \dfrac{\sqrt{3}  +  \sqrt{2} }{\sqrt{3}  +  \sqrt{2} }

Combine the fractions,

 \sf \implies   \dfrac{1(\sqrt{3}  +  \sqrt{2}) }{(\sqrt{3} -  \sqrt{2})(\sqrt{3}  +  \sqrt{2}) }

We know that,

 \sf \implies   (a - b)(a + b) = (a)^{2}  - (b)^{2}

So,

 \sf \implies   \dfrac{1(\sqrt{3}  +  \sqrt{2}) }{(\sqrt{3})^{2}  -  (\sqrt{2}) ^{2} }

 \sf \implies   \dfrac{1(\sqrt{3}  +  \sqrt{2}) }{3 -  2 }

 \sf \implies   \dfrac{1(\sqrt{3}  +  \sqrt{2}) }{1 }

 \sf \implies  ( \sqrt{3}  +  \sqrt{2})

Hence,

On rationalising we got,

\implies  \bf (\sqrt{3}  +  \sqrt{2})

━━━━━━━━━━━━━━━━━━━━━━━

 \sf(b) \: \:  \:  \dfrac{1}{ 2 \sqrt{5}   + \sqrt{3}  }

On rationalising,

   \sf \implies\dfrac{1}{ 2 \sqrt{5}   + \sqrt{3}  }  \times  \dfrac{2 \sqrt{5}    - \sqrt{3}}{2 \sqrt{5}    -  \sqrt{3}}

Combine the fractions,

   \sf \implies  \dfrac{1(2 \sqrt{5}    - \sqrt{3})}{(2 \sqrt{5}   + \sqrt{3})(2 \sqrt{5}    -  \sqrt{3})}

We know that,

 \sf \implies   (a  + b)(a  - b) = (a)^{2}  - (b)^{2}

So,

   \sf \implies  \dfrac{1(2 \sqrt{5}    - \sqrt{3})}{(2 \sqrt{5} )^{2}   -  (\sqrt{3})^{2} }

   \sf \implies  \dfrac{1(2 \sqrt{5}    - \sqrt{3})}{20 - 3 }

   \sf \implies  \dfrac{1(2 \sqrt{5}    - \sqrt{3})}{17 }

   \sf \implies  \dfrac{2 \sqrt{5}    - \sqrt{3}}{17 }

Hence,

On rationalising we got,

   \bf \implies\dfrac{2 \sqrt{5} - \sqrt{3}}{17 }


BrainlyPhantom: Awesome!
Similar questions