જો x = {{sqrt 5 + sqrt 2 } over {sqrt 5 - sqr | vedclass

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Question

(b) $y = {1 \over x}$ $ \Rightarrow $ $xy = 1$

$	herefore 3{x^2} + 4xy - 3{y^2} = 3\,(x - y)\,(x + y + 4)$

$ = 3\,.\,\left( {{{\sqrt 5 + \sqrt

Vedclass · Accepted Answer

${1 \over 3}[56\sqrt {10} + 12]$

Vedclass · Answer

(b) $y = {1 \over x}$ $ \Rightarrow $ $xy = 1$

$	herefore 3{x^2} + 4xy - 3{y^2} = 3\,(x - y)\,(x + y + 4)$

$ = 3\,.\,\left( {{{\sqrt 5 + \sqrt 2 } \over {\sqrt 5 - \sqrt 2 }} - {{\sqrt 5 - \sqrt 2 } \over {\sqrt 5 + \sqrt 2 }}} ight)\,\,\left( {{{\sqrt 5 + \sqrt 2 } \over {\sqrt 5 - \sqrt 2 }} + {{\sqrt 5 - \sqrt 2 } \over {\sqrt 5 + \sqrt 2 }}} ight)\, + 4$

$ = {{3\,[{{(\sqrt 5 + \sqrt 2 )}^2} - {{(\sqrt 5 - \sqrt 2 )}^2}]} \over {(5 - 2)\,(5 - 2)}}\,[{(\sqrt 5 + \sqrt 2 )^2} + {(\sqrt 5 - \sqrt 2 )^2}] + 4$

$ = {1 \over 3}.4\sqrt {10} \,.\,2\,(5 + 2) + 4 = {{56} \over 3}\sqrt {10} + 4 = {1 \over 3}(56\sqrt {10} + 12)$.

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} = $

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