Math, asked by kumariabha2341, 1 month ago

rationalise the denominator of each of the following

plz give the solution fast​

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Answers

Answered by ravichawan1977
1

Answer:

May I know which class are you?

Step-by-step explanation:

I don't know it so

Answered by mathdude500
4

\large\underline{\bf{Solution-}}

Answer :- i

 \red{\bf :\longmapsto\:\dfrac{4}{3 +  \sqrt{3} } }

Multiply and Divide by conjugate of denominator, we get

\:  \: \rm  =  \:  \: \dfrac{4}{3 +  \sqrt{3}} \times \dfrac{3 -  \sqrt{3} }{3 -  \sqrt{3} }

\:  \: \rm  =  \:  \: \dfrac{4(3 -  \sqrt{3})}{ {3}^{2} -  {( \sqrt{3}) }^{2}  }

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \boxed{ \sf \:  \because \: (x + y)(x - y) =  {x}^{2} -  {y}^{2}}

\:  \: \rm  =  \:  \: \dfrac{4(3 -  \sqrt{3})}{9 - 3}

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

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

 \:  \:  \:  \:  \:  \:  \: \red{ \boxed{\bf\implies \:\:  \dfrac{4}{3 -  \sqrt{3} }  =  \:  \: \dfrac{2(3 -  \sqrt{3})}{3}}}

Answer :- ii

 \green{\bf :\longmapsto\:\dfrac{5}{ \sqrt{2}  +  \sqrt{3} } }

\:  \: \rm  =  \:  \: \dfrac{5}{ \sqrt{3} +  \sqrt{2}  }

Multiply and Divide by conjugate of denominator, we get

\:  \: \rm  =  \:  \: \dfrac{5}{ \sqrt{3} +  \sqrt{2}}  \times \dfrac{ \sqrt{3}  -  \sqrt{2} }{ \sqrt{3}  -  \sqrt{2} }

\:  \: \rm  =  \:  \: \dfrac{5( \sqrt{3}  -  \sqrt{2}) }{ {( \sqrt{3}) }^{2} -  {( \sqrt{2} )}^{2}  }

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \boxed{ \sf \:  \because \: (x + y)(x - y) =  {x}^{2} -  {y}^{2}}

\:  \: \rm  =  \:  \: \dfrac{5( \sqrt{3} -  \sqrt{2})}{3 - 2}

\:  \: \rm  =  \:  \: 5( \sqrt{3} -  \sqrt{2})

 \:  \:  \:  \:  \:  \:  \: \green{ \boxed{\bf\implies \:\:  \dfrac{5}{ \sqrt{3}   +   \sqrt{2} }  =  \:  \: 5( \sqrt{3}  -  \sqrt{2})}}

Answer :- iii

 \blue{\bf :\longmapsto\:\dfrac{6 \sqrt{5} - 5 \sqrt{3}  }{2 \sqrt{3} + 4 \sqrt{5}}}

\:  \: \rm  =  \:  \: \dfrac{6 \sqrt{5} - 5 \sqrt{3}  }{4 \sqrt{5}  + 2 \sqrt{3} }

Multiply and Divide by conjugate of denominator, we get

\:  \: \rm  =  \:  \: \dfrac{6 \sqrt{5} - 5 \sqrt{3}  }{4 \sqrt{5}  + 2 \sqrt{3} } \times \dfrac{4 \sqrt{5}  - 2 \sqrt{3} }{4 \sqrt{5}  - 2 \sqrt{3} }

\:  \: \rm  =  \:  \: \dfrac{120 - 12 \sqrt{15} - 20 \sqrt{15}   + 30}{ {(4 \sqrt{5} )}^{2}  -  {(2 \sqrt{3} )}^{2} }

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \boxed{ \sf \:  \because \: (x + y)(x - y) =  {x}^{2} -  {y}^{2}}

\:  \: \rm  =  \:  \: \dfrac{150 - 32 \sqrt{15} }{80 - 12}

\:  \: \rm  =  \:  \: \dfrac{150 - 32 \sqrt{15} }{68}

\:  \: \rm  =  \:  \: \dfrac{75 - 16 \sqrt{15} }{34}

 \:  \:  \:  \:  \:  \: \blue{ \boxed{\bf :\implies\:\dfrac{6 \sqrt{5} - 5 \sqrt{3}  }{2 \sqrt{3} + 4 \sqrt{5}} = \dfrac{75 - 16 \sqrt{15} }{34} }}

Additional Information

More Identities to know:

  • (a + b)² = a² + 2ab + b²

  • (a - b)² = a² - 2ab + b²

  • a² - b² = (a + b)(a - b)

  • (a + b)² = (a - b)² + 4ab

  • (a - b)² = (a + b)² - 4ab

  • (a + b)² + (a - b)² = 2(a² + b²)

  • (a + b)³ = a³ + b³ + 3ab(a + b)

  • (a - b)³ = a³ - b³ - 3ab(a - b)
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