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Home  >>  EAMCET  >>  Mathematics
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$\sqrt {12-\sqrt {68+48 {\sqrt {2}}}}$ is equal to :

\[\begin {array} {1 1} (a)\;\sqrt 2-3 & \quad (b)\;2+\sqrt 2 \\ (c)\;2 - \sqrt 2 & \quad (d)\;6-2 \sqrt 8 \end {array}\]
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1 Answer

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$ (c)\;2 - \sqrt 2$
answered Nov 7, 2013 by pady_1
 
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