Question

Rationalise the denominator
$\boxed{\tt \dfrac{ (\sqrt{7 } - 1)}{( \sqrt{7} + 1) }}$
​

Answer 1

$\Huge \purple {\boxed {\underline {\mathbb {ANSWER:}}}}$

$\Huge \pink {\tt {\frac {(\sqrt{7 } - 1)}{( \sqrt{7} + 1) }}}$

$\huge \pink {\tt { \implies \frac {(\sqrt{7} - 1)}{(\sqrt{7}+1)} \times \frac{(\sqrt{7}-1)}{(\sqrt{7}-1)}}}$

$\huge \pink {\tt { \implies \frac {{(\sqrt{7}-1)}^{2} }{7-1}}}$

$\huge \pink {\tt { \implies \frac{8-2\sqrt{7}}{6}}}$

$\huge \pink {\tt { \implies \frac{2(4-\sqrt{7})}{6}}}$

$\huge \pink {\tt { \implies \frac{4-\sqrt{7}}{3}}}$

Answer 2

$\huge \text{ \blue{Question - }}$

$\longrightarrow{\tt \dfrac{ (\sqrt{7 } - 1)}{( \sqrt{7} + 1) }}$

$\huge \text{ \blue{Answer - }}$

$\longrightarrow{\tt \dfrac{ (\sqrt{7 } - 1)}{( \sqrt{7} + 1) }}$

First we will multiply numerator and denominator by √7-1

$\Longrightarrow{\tt \dfrac{ (\sqrt{7 } - 1)}{( \sqrt{7} + 1) } \times\dfrac{ (\sqrt{7 } - 1)}{( \sqrt{7} - 1) } }$

Identity we have to use :-

• Identity we have to use :-( a - b )² = ( a² + b² - 2ab )

• Identity we have to use :-( a - b )² = ( a² + b² - 2ab )a² - b² = ( a + b ) ( a - b )

$\Longrightarrow{\tt \dfrac{ (\sqrt{7 } - 1) ^{2} }{( \sqrt{7} ) ^{2} + (1)^{2}}}$

$\Longrightarrow{\tt \dfrac{ (\sqrt{7 }) ^{2} + ( 1) ^{2} - 2( \sqrt{7})(1) }{( \sqrt{7} ) ^{2} + (1)^{2}}}$

$\Longrightarrow{\tt \dfrac{7 + 1 - 2 \sqrt{7} }{7 - 1} }$

$\Longrightarrow{\tt \dfrac{8 - 2 \sqrt{7} }{6} }$

$\Longrightarrow{\tt \dfrac{ 2(4-\sqrt{7}) }{6} }$

$\Longrightarrow \red{{\tt \dfrac{ 2- \sqrt{7} }{3}} }$