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  1. #1
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    prove

    without a calculator

    prove
    \sqrt{2+\sqrt{3}}+\sqrt{4-\sqrt{7}}=\sqrt{5+\sqrt{21}}
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  2. #2
    Bar0n janvdl's Avatar
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    Quote Originally Posted by perash View Post
    without a calculator

    prove
    \sqrt{2+\sqrt{3}}+\sqrt{4-\sqrt{7}}=\sqrt{5+\sqrt{21}}
    You already posted this here
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  3. #3
    Super Member wingless's Avatar
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    Quote Originally Posted by janvdl View Post
    You already posted this here
    ?

    -----

    \sqrt{5+\sqrt{21}}=\sqrt{2+\sqrt{3}}+\sqrt{4-\sqrt{7}}

    5+\sqrt{21}=(\sqrt{2+\sqrt{3}}+\sqrt{4-\sqrt{7}})^2

    5+\sqrt{21}=2+\sqrt{3} + 4-\sqrt{7} + 2\sqrt{8 - 2\sqrt{7} + 4\sqrt{3} -\sqrt{21}}

    2\sqrt{8 - 2\sqrt{7} + 4\sqrt{3} -\sqrt{21}} = \sqrt{21} - \sqrt{3} + \sqrt{7} - 1

    8 - 2\sqrt{7} + 4\sqrt{3} -\sqrt{21} = (\frac{\sqrt{21} - \sqrt{3} + \sqrt{7} - 1}{2})^2

    ............ (a few long but easy calculations here..)

    8 - 2\sqrt{7} + 4\sqrt{3} -\sqrt{21} = 8 - 2\sqrt{7} + 4\sqrt{3} -\sqrt{21} \text{ } \blacksquare

    If you need help with the calculations I didn't show, just tell me
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  4. #4
    Bar0n janvdl's Avatar
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    Quote Originally Posted by wingless View Post
    ?

    -----
    He edited the post afterwards.
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  5. #5
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by perash View Post
    without a calculator

    prove
    \sqrt{2+\sqrt{3}}+\sqrt{4-\sqrt{7}}=\sqrt{5+\sqrt{21}}
    \sqrt {2 + \sqrt 3 }  = \frac{{\sqrt {4 + 2\sqrt 3 } }}<br />
{{\sqrt 2 }} = \frac{{\sqrt {\left( {1 + \sqrt 3 } \right)^2 } }}<br />
{{\sqrt 2 }} = \frac{{1 + \sqrt 3 }}<br />
{{\sqrt 2 }}.

    In the same way \sqrt {4 - \sqrt 7 }  = \frac{{\sqrt 7  - 1}}<br />
{{\sqrt 2 }}.

    Finally \sqrt {2 + \sqrt 3 }  + \sqrt {4 - \sqrt 7 }  = \frac{{\sqrt 3  + \sqrt 7 }}<br />
{{\sqrt 2 }} = \frac{{\sqrt {10 + 2\sqrt {21} } }}<br />
{{\sqrt 2 }} = \sqrt {5 + \sqrt {21} } \quad\blacksquare
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