The rationalising factor of $2\sqrt 3 - \sqrt 7 $ is
The rationalising factor of $2\sqrt 3 - \sqrt 7 $ is
- A
$\sqrt 3 + \sqrt 7 $
- B
$2\sqrt 3 + \sqrt 7 $
- C
$\sqrt 3 + 2\sqrt 7 $
- D
None of these
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${{\sqrt {(5/2)} + \sqrt {(7 - 3\sqrt 5 )} } \over {\sqrt {(7/2)} + \sqrt {(16 - 5\sqrt 7 )} }}=$
${{\sqrt {(5/2)} + \sqrt {(7 - 3\sqrt 5 )} } \over {\sqrt {(7/2)} + \sqrt {(16 - 5\sqrt 7 )} }}=$
Solution of the equation ${9^x} - {2^{x + {1 \over 2}}} = {2^{x + {3 \over 2}}} - {3^{2x - 1}}$
Solution of the equation ${9^x} - {2^{x + {1 \over 2}}} = {2^{x + {3 \over 2}}} - {3^{2x - 1}}$
$\root 4 \of {(17 + 12\sqrt 2 )} = $
$\root 4 \of {(17 + 12\sqrt 2 )} = $
$\sqrt {(3 + \sqrt 5 )} $ is equal to
$\sqrt {(3 + \sqrt 5 )} $ is equal to
The greatest number among $\root 3 \of 9 ,\root 4 \of {11} ,\root 6 \of {17} $ is
The greatest number among $\root 3 \of 9 ,\root 4 \of {11} ,\root 6 \of {17} $ is