Math, asked by bada11, 1 year ago

simplify: (√5+√3/√5-√3)+(√2+√3/√2-√3)

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Answers

Answered by BrainlyPopularman
6

TO SOLVE :

  \\{ \bold{ \dfrac{ \sqrt{5}  +  \sqrt{3} }{ \sqrt{5} -  \sqrt{3}  }  +  \dfrac{ \sqrt{2}  +  \sqrt{3} }{ \sqrt{2}  -  \sqrt{3} }  }}  \\

SOLUTION :

Let the function be –

  \\ \implies{ \bold{y =  \dfrac{ \sqrt{5}  +  \sqrt{3} }{ \sqrt{5} -  \sqrt{3}  }  +  \dfrac{ \sqrt{2}  +  \sqrt{3} }{ \sqrt{2}  -  \sqrt{3} }  }}  \\

• We should write this as –

  \\ \implies{ \bold{y = p +  q}}  \\

▪︎ Now first solve 'p' –

  \\ \implies{ \bold{p =\dfrac{ \sqrt{5}  +  \sqrt{3} }{ \sqrt{5} -  \sqrt{3}}}}  \\

• Now rationalization –

  \\ \implies{ \bold{p =\dfrac{ \sqrt{5}  +  \sqrt{3} }{ \sqrt{5} -  \sqrt{3}} \times  \dfrac{ \sqrt{5}  +  \sqrt{3} }{ \sqrt{5}  +  \sqrt{3} } }}  \\

  \\ \implies{ \bold{p =\dfrac{ (\sqrt{5}  +  \sqrt{3} )^{2} }{ (\sqrt{5} )^{2} -  (\sqrt{3})^{2} }}}  \\

  \\ \implies{ \bold{p =\dfrac{5 + 3 + 2 \sqrt{5}  \sqrt{3} }{5 - 3}}}  \\

  \\ \implies{ \bold{p =\dfrac{8 + 2 \sqrt{5}  \sqrt{3} }{2}}}  \\

  \\ \implies{ \bold{p =\dfrac{2(4 +  \sqrt{5}  \sqrt{3} )}{2}}}  \\

  \\ \implies \large{ \boxed{ \bold{p =4+ \sqrt{15} }}}  \\

▪︎ Now solving 'q' –

  \\ \implies{ \bold{q = \dfrac{ \sqrt{2}  +  \sqrt{3} }{ \sqrt{2}  -  \sqrt{3} } }}  \\

• Now rationalization –

  \\ \implies{ \bold{q = \dfrac{ \sqrt{2}  +  \sqrt{3} }{ \sqrt{2}  -  \sqrt{3} } \times  \dfrac{ \sqrt{2}  +  \sqrt{3} }{ \sqrt{2}   +   \sqrt{3} }  }}  \\

  \\ \implies{ \bold{q = \dfrac{ (\sqrt{2}  +  \sqrt{3} )^{2} }{ (\sqrt{2})^{2}   -  (\sqrt{3} )^{2} }}}  \\

  \\ \implies{ \bold{q = \dfrac{2 + 3 + 2 \sqrt{2} \sqrt{3}  }{2 - 3}}}  \\

  \\ \implies{ \bold{q = \dfrac{5+ 2 \sqrt{2} \sqrt{3}  }{ - 1}}}  \\

  \\ \implies \large{ \boxed{ \bold{q = - 5 - 2 \sqrt{6} }}}  \\

• So that –

  \\ \implies{ \bold{y =4+ \sqrt{15}- 5 - 2 \sqrt{6}}}  \\

  \\ \implies \large{ \boxed{{ \bold{y =  \sqrt{15} - 2 \sqrt{6} - 1}}}}  \\

Answered by MaIeficent
22

Step-by-step explanation:

{\red{\underline{\underline{\bold{Given:-}}}}}

  • \frac{ \sqrt{5 } +  \sqrt{3}  }{ \sqrt{5} - 3 }  +  \frac{ \sqrt{2} +  \sqrt{3}  }{ \sqrt{2}  -  \sqrt{3} }

{\blue{\underline{\underline{\bold{Concept\:used\:here}}}}}

  • Rationalising of the numbers.

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

\frac{ \sqrt{5 } +  \sqrt{3}  }{ \sqrt{5} - 3 }  +  \frac{ \sqrt{2} +  \sqrt{3}  }{ \sqrt{2}  -  \sqrt{3} }

______________

\frac{ \sqrt{5} +  \sqrt{3}  }{ \sqrt{5}  -  \sqrt{3} }

By rationalising:-

 \frac{ \sqrt{5} +  \sqrt{3}  }{ \sqrt{5}  -  \sqrt{3} }  \times  \frac{ \sqrt{5}  +  \sqrt{3} }{ \sqrt{5} +  \sqrt{3}  }  \\  \\  =    \frac{{( \sqrt{5} +  \sqrt{3})  }^{2}}{ {( \sqrt{5} )}^{2} -  {( \sqrt{3}) }^{2}  }   \\  \\  =  \frac{ {( \sqrt{5}) }^{2}  +  {( \sqrt{3} )}^{2}  + 2( \sqrt{5}) (\sqrt{3}  )}{5 - 3}  \\  \\  =  \frac{5 + 3 + 2 \sqrt{15} }{2}  \\  \\  =  \frac{8 + 2 \sqrt{15} }{2}  \\  \\  =  \frac{2(4 +  \sqrt{15}) }{2}  \\  \\  = 4 +  \sqrt{15}

_______________

\frac{ \sqrt{2} +  \sqrt{3}  }{ \sqrt{2}  -  \sqrt{3} }  \times  \frac{ \sqrt{2}  +  \sqrt{3} }{ \sqrt{2} +  \sqrt{3}  }  \\  \\  =    \frac{{( \sqrt{2} +  \sqrt{3})  }^{2}}{ {( \sqrt{2} )}^{2} -  {( \sqrt{3}) }^{2}  }   \\  \\  =  \frac{ {( \sqrt{2}) }^{2}  +  {( \sqrt{3} )}^{2}  + 2( \sqrt{2}) (\sqrt{3}  )}{2 - 3}  \\  \\  =  \frac{2 + 3 + 2 \sqrt{6} }{ - 1}  \\  \\  =  \frac{5 + 2 \sqrt{6} }{ - 1} \\ \\ =  - 5 - 2 \sqrt{6}

_______________

\frac{ \sqrt{5 } +  \sqrt{3}  }{ \sqrt{5} - 3 }  +  \frac{ \sqrt{2} +  \sqrt{3}  }{ \sqrt{2}  -  \sqrt{3} }

4 +  \sqrt{15}  - 5 - 2 \sqrt{6}

\boxed{ = \sqrt{15}   - 2 \sqrt{6} -1 }

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