Math, asked by kslp1234, 5 months ago

Simplify the following

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Answered by aryan073
6

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\huge\sf\green{Answer}

 \:  \boxed { \rm \pink{simplify \:  \: the \:  \: following}}

 \implies \bf \red{ \frac{1}{1 +  \sqrt{2} }  +  \frac{1}{ \sqrt{2} +  \sqrt{3}  }  +  \frac{1}{ \sqrt{3} +  \sqrt{4}  } }

 \:  \dashrightarrow \bf{ \frac{ \sqrt{2 }  +  \sqrt{3}  + 1 +  \sqrt{2} }{1( \sqrt{2} +  \sqrt{3} ) +  \sqrt{2} ( \sqrt{2} +  \sqrt{3}   }  +  \frac{1}{ \sqrt{3}  +  \sqrt{4} } }

 \:  \dashrightarrow \bf{ \frac{2 \sqrt{2}  +  \sqrt{3}  + 1}{ \sqrt{2}  +  \sqrt{3} + 2 +  \sqrt{6}  }  +  \frac{1}{ \sqrt{3}  +  \sqrt{4} } }

 \:  \dashrightarrow \bf{ \frac{ \sqrt{3}( \sqrt{2}   +  \sqrt{3}  + 1 +  \sqrt{2} ) +  \sqrt{4} ( \sqrt{2} +  \sqrt{3}  + 1 +  \sqrt{2} )  +  \sqrt{2} +  \sqrt{3}  + 2 +  \sqrt{6}  }{ \sqrt{3} ( \sqrt{2}  +  \sqrt{3}  +  2 +  \sqrt{6} ) +  \sqrt{4}( \sqrt{2} +  \sqrt{3}    + 2 +  \sqrt{6}  }  }

 \dashrightarrow \bf{ \frac{ \sqrt{6} + 3 +  \sqrt{3}   +  \sqrt{6} +  \sqrt{8} +  \sqrt{12}   +  \sqrt{4}   +  \sqrt{8}  +  \sqrt{2}  +  \sqrt{3}  + 2 \sqrt{2} +  \sqrt{12}  }{ \sqrt{6} + 3 + 2 \sqrt{3} +  \sqrt{18}   +  \sqrt{8}  +  \sqrt{12}  + 2 \sqrt{4}  +  \sqrt{24}  } }

 \dashrightarrow \bf{ \frac{2  \sqrt{6} + 3 +  \sqrt{3}  +  2 \sqrt{2} + 2 \sqrt{3}   + 2 + 2 \sqrt{2}  +  \sqrt{2}  +  \sqrt{3} + 2 \sqrt{2}  + 2 \sqrt{3}   }{ \sqrt{6} + 3 + 2 \sqrt{3}   + 3 \sqrt{2} + 2 \sqrt{2}  + 2 \sqrt{3}   + 2 \times 2 + 2 \sqrt{6} } }

 \dashrightarrow \bf{ \frac{2 \sqrt{6} + 5 + 3 \sqrt{3}  + 5 \sqrt{2}  + 3 \sqrt{3}  + 2 \sqrt{2}  }{ \sqrt{6}  + 7 + 4 \sqrt{3} + 5 \sqrt{2}  + 2 \sqrt{6}}  }

 \pink \bigstar \boxed { \bf \purple{  \implies\frac{2 \sqrt{ 6}  + 5 + 7\sqrt2+ 6 \sqrt{3}  }{ 3 \sqrt{6}  + 7 + 4 \sqrt{3} + 5 \sqrt{2}  } }}

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