Question
यदि $x = \sqrt {7 + 4\sqrt 3 } ,$ तब $x + \frac{1}{x} = $
$\therefore$ $\frac{1}{x} = \frac{1}{{\sqrt {7 + 4\sqrt 3 } }} = \frac{{\sqrt {7 - 4\sqrt 3 } }}{{\sqrt {7 + 4\sqrt 3 } .\sqrt {7 - 4\sqrt 3 } }}$
$ = \sqrt {7 - 4\sqrt 3 } $
$\therefore$ $x + \frac{1}{x} = \sqrt {7 + 4\sqrt 3 } + \sqrt {7 - 4\sqrt 3 } $
$ = (\sqrt 3 + 2) + (2 - \sqrt 3 ) = 4$
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$a R b \Leftrightarrow a, b^2$ को विभाजित करता है.
$I$. सतुल्यता $(reflexivity)$
$II$. सममिति $(symmetry)$
$III$. संक्रमिता $(transitivity)$