Math, asked by shuvam1644, 2 days ago

plz answer the question​

Attachments:

Answers

Answered by royarpita639
1

Step-by-step explanation:

hope this helps you and look the answer once

Attachments:
Answered by Anonymous
38

Answer:

\mathtt\purple{❖SOLUTION:-}

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:

=\frac{1}{1 +  \sqrt{2} }    +  \frac{1}{\sqrt{2} +  \sqrt{3}  }   +  \frac{1}{ \sqrt{3} +  \sqrt{4}  }   +   \frac{1}{ \sqrt{4}  +  \sqrt{5} } +  \frac{1}{ \sqrt{5}   +   \sqrt{6} }  +  \frac{1}{ \sqrt{6}  +  \sqrt{7} } +  \frac{1}{ \sqrt{7}   +  \sqrt{8} }  +  \frac{1}{ \sqrt{8} +  \sqrt{9} }

 = \frac{1}{1 +  \sqrt{2} } \times  \frac{ 1-  \sqrt{2}  }{1 -  \sqrt{2} }   +  \frac{1}{\sqrt{2} +  \sqrt{3}  }  \times  \frac{ \sqrt{2} -  \sqrt{3}  }{ \sqrt{2}  -  \sqrt{3} }  +  \frac{1}{ \sqrt{3} +  \sqrt{4}  }  \times  \frac{ \sqrt{3}  -  \sqrt{4} }{ \sqrt{3} -  \sqrt{4}  }  +   \frac{1}{ \sqrt{4}  +  \sqrt{5} }  \times  \frac{ \sqrt{4}  -  \sqrt{5} }{ \sqrt{4} -  \sqrt{5}  }  +  \frac{1}{ \sqrt{5}   +   \sqrt{6} }  \times   \frac{ \sqrt{5}  -  \sqrt{6} }{ \sqrt{5}  -  \sqrt{6} }  +  \frac{1}{ \sqrt{6}  +  \sqrt{7} } \times  \frac{ \sqrt{6}  +  \sqrt{7} }{ \sqrt{6}  -  \sqrt{7} } +  \frac{1}{ \sqrt{7}   +  \sqrt{8} }  \times    \frac{ \sqrt{7}  -  \sqrt{8}  }{ \sqrt{7}  -  \sqrt{8} }  +  \frac{1}{ \sqrt{8} +  \sqrt{9}  }  \times  \frac{ \sqrt{8}   -  \sqrt{9} }{ \sqrt{8}  -  \sqrt{9} }

=\frac{1 -  \sqrt{2}  }{ {1}^{2}  -  { (\sqrt{2}) }^{2} }  +  \frac{ \sqrt{2} -  \sqrt{3}  }{ { (\sqrt{2} )}^{2}  -  {( \sqrt{3}) }^{2} }  +  \frac{ \sqrt{3}  -  \sqrt{4} }{ {( \sqrt{3}) }^{2}  - ( \sqrt{4}) ^{2}  }  +  \frac{ \sqrt{4}  -  \sqrt{5} }{ { (\sqrt{4} )}^{2}  -  { (\sqrt{5} )}^{2} }  +  \frac{ \sqrt{5}  -  \sqrt{6} }{ {( \sqrt{5} )}^{2}  -  (\sqrt{6}  )^{2} }  +  \frac{ \sqrt{6}  -  \sqrt{7}  }{ {( \sqrt{6} )}^{2}  +  {( \sqrt{7} )}^{2} }  +  \frac{ \sqrt{7}  -  \sqrt{8} }{ ({ \sqrt{7} )}^{2} -  { (\sqrt{8} )}^{2} }  +  \frac{ \sqrt{8}  -  \sqrt{9} }{ {( \sqrt{8}) }^{2}  -  { (\sqrt{9}) }^{2} }

 =  \frac{1 -  \sqrt{2}  }{  - 1 }  +  \frac{ \sqrt{2} -  \sqrt{3}  }{  - 1}  +  \frac{ \sqrt{3}  -  \sqrt{4} }{  - 1}  +  \frac{ \sqrt{4}  -  \sqrt{5} }{ - 1}  +  \frac{ \sqrt{5}  -  \sqrt{6} }{ - 1 }  +  \frac{ \sqrt{6}  -  \sqrt{7}  }{  - 1}  +  \frac{ \sqrt{7}  -  \sqrt{8} }{  - 1}  +  \frac{ \sqrt{8}  -  \sqrt{9} }{  - 1}

  = - (1 -  \sqrt{2} ) - ( \sqrt{2}  -  \sqrt{3} ) - ( \sqrt{3}  -  \sqrt{4} ) - ( \sqrt{4}  -  \sqrt{5} ) - ( \sqrt{5}  -  \sqrt{6} ) - ( \sqrt{6}  -  \sqrt{7} ) - ( \sqrt{7}  - \sqrt{8} ) - ( \sqrt{8}  -  \sqrt{9} )

 =  - 1 +  \sqrt{2 }  -  \sqrt{2}  +  \sqrt{3}  -  \sqrt{3}  +  \sqrt{4}  -  \sqrt{4}  +  \sqrt{5}  -  \sqrt{5}  +  \sqrt{6}  -  \sqrt{6}    +   \sqrt{7}     -  \sqrt{7}   + \sqrt{8}   -  \sqrt{8}   +   \sqrt{9}

 =  - 1 +  \sqrt{9}

 =  - 1 +  \sqrt{3 \times3}

 =  - 1 + 3

 = 2

\mathtt\purple{HENCE \: , VERIFIED}

Similar questions