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Akshit_math

  • 2 years ago

The value of _____

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  1. hartnn
    • 2 years ago
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    is _______

  2. Akshit_math
    • 2 years ago
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    \[\sqrt{1 + 2008\sqrt{1 + 2009\sqrt{1 + 2010\sqrt{1 + 2011.2013}}}}\]

  3. Akshit_math
    • 2 years ago
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    what is the value of the above ???

  4. Mimi_x3
    • 2 years ago
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    do you have access to a calculator? or do you want to solve it by hand?

  5. Akshit_math
    • 2 years ago
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    we are Indians and here most of the time its by hand ...so please try to solve and help me out without the use of calculator,,

  6. hartnn
    • 2 years ago
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    ok, now can u write 2011*2013 as (2012-1)(2012+1) ?? and do u know what (a+b)(a-b) =??

  7. Akshit_math
    • 2 years ago
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    yeah \[(a+b)(a-b)=a ^{2}-b ^{2}\]

  8. hartnn
    • 2 years ago
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    good, so what will be (2012-1)(2012+1) =??

  9. Akshit_math
    • 2 years ago
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    \[2012^{2}-1\]

  10. hartnn
    • 2 years ago
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    and u had 1+ 2011*2013 so now whats 1+2012^2-1 ??

  11. Akshit_math
    • 2 years ago
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    heyyyyy thaaaanzzznnzzzz buddy i got it all after dat ...u made it so easy 4 me now!!!...champ!!!..thnak u!

  12. hartnn
    • 2 years ago
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    welcome :)

  13. Akshit_math
    • 2 years ago
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    hey @sriramkumar u typing since long tym eh!

  14. sriramkumar
    • 2 years ago
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    let 2012=a, then the problemn becomes \[\sqrt{1+(a-4)\sqrt{1+(a-3)\sqrt{1+(a-2)\sqrt{1+(a-1)(a+1)}}}}\] then the inner root becomes \[a ^{2}-1 +1=a ^{2}\] \[1+(a-2)\sqrt{a ^{2}}=1+(a-2)a=1+a ^{2}-2a=(a-1)^{2}\] ...... and so on....

  15. Akshit_math
    • 2 years ago
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    damn gud!!!....@sriramkumar

  16. Akshit_math
    • 2 years ago
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    n the answer is 2009

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