Math, asked by krishkothari72, 4 months ago

rationalise the denominator​

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Answered by jaimindave
0

Hope It's Helpful.

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Answered by saanvigrover2007
18

 \sf \green{Question : } \:  \sf \red{ \frac{ \sqrt{7}  +  \sqrt{3} }{ \sqrt{7}  -  \sqrt{2} } } \\

 \sf \green{Solution : }

 \sf{ \twoheadrightarrow  \frac{\sqrt{7}  +  \sqrt{3}}{ \sqrt{7}  -  \sqrt{2} }  \times  \frac{ \sqrt{7} +  \sqrt{2}  }{ \sqrt{7} +  \sqrt{2}  } } \\

 \sf{ \twoheadrightarrow \frac{( \sqrt{7} +  \sqrt{3} )( \sqrt{7} +  \sqrt{2} )  }{ { (\sqrt{7} )}^{2}  -  { (\sqrt{2}) }^{2} } } \\

 { \rm{ \blue{《{a}^{2}  -  {b}^{2} = (a - b)(a + b) 》}}}

 \sf{  \twoheadrightarrow\frac{ \sqrt{7}( \sqrt{7} +  \sqrt{2}  ) +  \sqrt{3}( \sqrt{7}  +  \sqrt{2} )}{7 - 2}  } \\

 \sf{  \twoheadrightarrow\frac{7 +  \sqrt{7} \sqrt{2} +  \sqrt{3} \sqrt{7}  +  \sqrt{3} \sqrt{2}    }{5} }  \\

  \sf{{ \pink{\twoheadrightarrow  \frac{7 +  \sqrt{14} +  \sqrt{21}  +  \sqrt{6}  }{5} }}} \\

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