{{sqrt 2 } over {sqrt {(2 + sqrt 3 )} - sqrt | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(b) ${{\sqrt 2 } \over {\sqrt {2 + \sqrt 3 } - \sqrt {2 - \sqrt 3 } }} = {{\sqrt 2 \,\left( {\sqrt {2 + \sqrt 3 } + \sqrt {2 - \sqrt 3 } } \right)} \o

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(b) ${{\sqrt 2 } \over {\sqrt {2 + \sqrt 3 } - \sqrt {2 - \sqrt 3 } }} = {{\sqrt 2 \,\left( {\sqrt {2 + \sqrt 3 } + \sqrt {2 - \sqrt 3 } } \right)} \over {(2 + \sqrt 3 ) - (2 - \sqrt 3 )}}$

$ = {{\sqrt {4 + 2\sqrt 3 } + \sqrt {4 - 2\sqrt 3 } } \over {2\sqrt 3 }} = {{(\sqrt 3 + 1)\, + \,(\sqrt 3 - 1)} \over {2\sqrt 3 }} = 1$.

${{\sqrt 2 } \over {\sqrt {(2 + \sqrt 3 )} - \sqrt {(2 - \sqrt 3 } )}} = $

Similar Questions

${{\sqrt {(5/2)} + \sqrt {(7 - 3\sqrt 5 )} } \over {\sqrt {(7/2)} + \sqrt {(16 - 5\sqrt 7 )} }}=$

જો ${\left( {{2 \over 3}} \right)^{x + 2}} = {\left( {{3 \over 2}} \right)^{2 - 2x}},$ તો $x =$

જો ${a^{x - 1}} = bc,{b^{y - 1}} = ca,{c^{z - 1}} = ab,$ તો $\sum {(1/x) = } $

$\sqrt {[12\sqrt 5 + 2\sqrt {(55)} ]} $ નું વર્ગમૂળ મેળવો.

સમીકરણ $\sqrt {(x + 1)} - \sqrt {(x - 1)} = \sqrt {(4x - 1)} $, $x \in R$ ને .. . .