Question

$rationalise \: the \: denominator \\ \frac{1}{ \sqrt{5 + \sqrt{2} } }$
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Answer 1

$\sf\huge\orange{\underbrace{ Question : }}$

Rationalise the denominator :

$\tt \cfrac{1}{\sqrt{5} + \sqrt{2}}$

$\sf\huge\red{\underbrace{ Solution : }}$

$\sf \implies \cfrac{1}{\sqrt{5} + \sqrt{2}}$

• Multiply both numerator and denominator with √5 - √2.

$\sf \implies \cfrac{1}{\sqrt{5} + \sqrt{2}} \times \cfrac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}$

$\sf \implies \cfrac{\sqrt{5}-\sqrt{2}}{(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})}$

• (a + b)(a - b) = a² - b²

$\sf \implies \cfrac{\sqrt{5}-\sqrt{2}}{(\sqrt{5})^{2} - (\sqrt{2})^{2}}$

$\sf \implies \cfrac{\sqrt{5}-\sqrt{2}}{5-2}$

$\sf \implies \cfrac{\sqrt{5}-\sqrt{2}}{3}$

$\underline{\boxed{\rm{\purple{\therefore \cfrac{1}{\sqrt{5} + \sqrt{2}} = \cfrac{\sqrt{5}-\sqrt{2}}{3} .}}}}\:\orange{\bigstar}$

Answer 2

ANSWER✔

$\large\underline\bold{GIVEN,}$

$\rm\dashrightarrow \dfrac{1}{ \sqrt{5 + \sqrt{2} } }$

$\large\underline\bold{TO\:FIND,}$

$\sf\dashrightarrow TO\:RATIONALISE\:THE\: DENOMINATOR$

IDENTITY IN USE,

$\large{\boxed{\bf{ (a + b)(a - b) = a^2 - b^2}}}$

$\large\underline\bold{SOLUTION,}$

$\sf\therefore \dfrac{1}{ \sqrt{5 + \sqrt{2} } }$

$\sf\dashrightarrow multiplying\:numerator\:and\:denomintaor\:by, \: \sqrt{5}- \sqrt{2}$

WE GET,

$\sf \implies \dfrac{1}{\sqrt{5} + \sqrt{2}} \times \dfrac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}$

$\sf \implies \dfrac{\sqrt{5}-\sqrt{2}}{(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})}$

$\sf \implies \dfrac{\sqrt{5}-\sqrt{2}}{(\sqrt{5})^{2} - (\sqrt{2})^{2}}$

$\sf \implies \dfrac{\sqrt{5}-\sqrt{2}}{(\sqrt{25}) - (\sqrt{4})}$

$\sf \implies \dfrac{\sqrt{5}-\sqrt{2}}{5-2}$

$\sf \implies \dfrac{\sqrt{5}-\sqrt{2}}{3}$

$\large{\boxed{\bf{\star\:\: \dfrac{1}{\sqrt{5} + \sqrt{2}} = \dfrac{\sqrt{5}-\sqrt{2}}{3} \:\: \star}}}$

HENCE THE RATIONALIST DENOMINATOR IS $\sf \dfrac{\sqrt{5}-\sqrt{2}}{3}$

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