Question

rationalise the denominator
,1/√7+√6-√13​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

$\qquad$❏$\tt\red{{ \dfrac{1}{ \sqrt{7} + \sqrt{6} - \sqrt{13} } }}$

Let:-

$\qquad$☀️ $\tt { \sqrt{7} + \sqrt{6} = a }$

$\qquad$☀️ $\tt { \sqrt{13} = b }$

$\qquad$❏ To rationalise the denominator, we'll have to multiply the denominator and the numerator by its rationalising factor. As we supposed (√7 + √6) as a and √13 as b, so the denominator is like ( a - b ). We know that rationalising factor of $\sf{ (a -b)}$ is $\sf { (a + b)}$. Therefore, the rationalising factor of $\bigg ( \sf { \dfrac{1}{\sqrt{7} + \sqrt{6} - \sqrt{13} } } \bigg )$ is $( \sf { \sqrt{7} + \sqrt{6} + \sqrt{13} } )$.

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$\qquad\leadsto\quad \tt { \dfrac{1}{ \sqrt{7} + \sqrt{6} - \sqrt{13} } \times \dfrac{ \sqrt{7} + \sqrt{6} + \sqrt{13} }{ \sqrt{7} + \sqrt{6} + \sqrt{13} } }$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\qquad\leadsto\quad \tt { \dfrac{ \sqrt{7} + \sqrt{6} + \sqrt{13} }{ { \big( \sqrt{7} + \sqrt{6} \big ) }^{2} - {( \sqrt{ 13} )}^{2} } }$

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$\qquad$❏(a - b)(a + b) = a² - b²

$\qquad\leadsto\quad \tt { \dfrac{ \sqrt{7} + \sqrt{6} + \sqrt{13} }{ {( \sqrt{7}) }^{2} + { (\sqrt{6}) }^{2} + 2( \sqrt{7} \times \sqrt{6} ) } - {( \sqrt{13}) }^{2} }$

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$\qquad\leadsto\quad \tt { \dfrac{ \sqrt{7} + \sqrt{6} + \sqrt{13} }{ 7 + 6 + 2( \sqrt{42} ) - 13 } }$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\qquad\leadsto\quad \tt { \dfrac{ \sqrt{7} + \sqrt{6} + \sqrt{13} }{ \cancel{ 13} + 2( \sqrt{42} ) \cancel{- 13} }}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\qquad\leadsto\quad \tt { \dfrac{ \sqrt{7} + \sqrt{6} + \sqrt{13} }{ 2( \sqrt{42} ) }}$

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$\qquad\leadsto\quad \tt { \dfrac{ \sqrt{7} + \sqrt{6} + \sqrt{13} }{ 2( \sqrt{42} ) } \times \dfrac{ \sqrt{42} }{ \sqrt{42} } }$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\qquad\leadsto\quad \tt { \dfrac{ \sqrt{42} (\sqrt{7} )+ \sqrt{42} (\sqrt{6} )+ \sqrt{42(} \sqrt{13}) }{ \sqrt{42} (2 \sqrt{42}) } }$

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$\qquad\leadsto\quad \tt { \dfrac{ \sqrt{294} + \sqrt{252} + \sqrt{546} }{ 2 \times 42 } }$

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$\qquad\leadsto\quad \tt { \dfrac{ \sqrt{294} + \sqrt{252} + \sqrt{546} }{ 84} }$

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$\qquad\leadsto\quad \tt { \dfrac{ 7\sqrt{6} + 6\sqrt{7} + \sqrt{546} }{ 84} }$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\qquad\leadsto\quad \tt { \dfrac{7\sqrt{6}}{84} + \dfrac{6\sqrt{7} }{84} + \dfrac{\sqrt{546} }{ 84} }$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\qquad\leadsto\quad \tt \red{\boxed{ \sf { \dfrac{\sqrt{6}}{12} + \dfrac{\sqrt{7} }{14} + \dfrac{\sqrt{546} }{ 84} }} }$