MCQ
${{12} \over {3 + \sqrt 5 - 2\sqrt 2 }} = $
- A$1 + \sqrt 5 + \sqrt {(10)} + \sqrt 2 $
- B$1 + \sqrt 5 - \sqrt {(10)} + \sqrt 2 $
- ✓$1 + \sqrt 5 + \sqrt {10} - \sqrt 2 $
- D$1 - \sqrt 5 - \sqrt 2 + \sqrt {(10)} $
$ = {{12\,(3 - 2\sqrt 2 - \sqrt 5 )} \over {{{(3 - 2\sqrt 2 )}^2} - 5}} = {{12\,(3 - 2\sqrt 2 - \sqrt 5 )} \over {17 - 12\sqrt 2 - 5}}$
$ = {{(3 - 2\sqrt 2 - \sqrt 5 )} \over {1 - \sqrt 2 }} = {{(\sqrt 5 + 2\sqrt 2 - 3)\,(\sqrt 2 + 1)} \over {(\sqrt 2 - 1)\,(\sqrt 2 + 1)}}$
$ = {{\sqrt {10} + 4 - 3\sqrt 2 + \sqrt 5 + 2\sqrt 2 - 3} \over {2 - 1}} = 1 + \sqrt 5 + \sqrt {10} - \sqrt 2 $.
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$\left| {1 - {{\log }_{\frac{1}{6}}}x} \right| + \left| {{{\log }_2}x} \right| + 2 = \left| {3 - {{\log }_{\frac{1}{6}}}x + {{\log }_{\frac{1}{2}}}x} \right|$ is $\left[ {\frac{a}{b},a} \right],a,b, \in N,$ then the value of $(a + b)$ is