$\sqrt {(3 + \sqrt 5 )} $ is equal to
$\sqrt {(3 + \sqrt 5 )} $ is equal to
- A
$\sqrt 5 + 1$
- B
$\sqrt 3 + \sqrt 2 $
- C
$(\sqrt 5 + 1)/\sqrt 2 $
- D
${1 \over 2}(\sqrt 5 + 1)$
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${4 \over {1 + \sqrt 2 - \sqrt 3 }} = $
${4 \over {1 + \sqrt 2 - \sqrt 3 }} = $
${{3\sqrt 2 } \over {\sqrt 6 + \sqrt 3 }} - {{4\sqrt 3 } \over {\sqrt 6 + \sqrt 2 }} + {{\sqrt 6 } \over {\sqrt 3 + \sqrt 2 }} = $
${{3\sqrt 2 } \over {\sqrt 6 + \sqrt 3 }} - {{4\sqrt 3 } \over {\sqrt 6 + \sqrt 2 }} + {{\sqrt 6 } \over {\sqrt 3 + \sqrt 2 }} = $
$\sqrt {(3 + \sqrt 5 )} - \sqrt {(2 + \sqrt 3 )} = $
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The rationalising factor of $2\sqrt 3 - \sqrt 7 $ is
The rationalising factor of $2\sqrt 3 - \sqrt 7 $ is