{{√ {(5/2)} + √ {(7 - 3√ 5 )} } over {√ {(7/2)} + √ {(16

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Gujarati

Hindi

Basic of Logarithms

medium

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

સંમેય

અમૂલદ

$\sqrt 7 $ નો ગુણક

એકપણ નહીં

Solution

(a) ${{\sqrt {5/2} + \sqrt {7 – 3\sqrt 5 } } \over {\sqrt {7/2} + \sqrt {16 – 5\sqrt 7 } }} = {{\sqrt 5 + \sqrt {14 – 6\sqrt 5 } } \over {\sqrt 7 + \sqrt {32 – 10\sqrt 7 } }}$

$= {{\sqrt 5 + (3 – \sqrt {5)} } \over {\sqrt 7 + (5 – \sqrt 7 )}} = {3 \over 5}$, which is rational.

Standard 11

Mathematics

જો $x = {2^{1/3}} – {2^{ – 1/3}},$ તો $2{x^3} + 6x = $

medium

જો ${a^x} = bc,{b^y} = ca,\,{c^z} = ab,$ તો $xyz=$

easy

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

easy

જો $x = {{\sqrt 5 + \sqrt 2 } \over {\sqrt 5 – \sqrt 2 }},y = {{\sqrt 5 – \sqrt 2 } \over {\sqrt 5 + \sqrt 2 }},$ તો $3{x^2} + 4xy – 3{y^2} = $

hard

જો $x = 3 – \sqrt {5,} $ તો ${{\sqrt x } \over {\sqrt 2 + \sqrt {(3x – 2)} }} = $