Question

simplify
$\frac{ \sqrt{7} - 1 }{ \sqrt{7} + 1} - \frac{ \sqrt{7} + 1}{ \sqrt{7 } - 1 } = a + b \sqrt{7}$
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Answer 1

$\large{ \bf\pink { \underline { \underline{Answer : }}}}$

${ \sf {\dfrac{ \sqrt{7} - 1}{ \sqrt{7} + 1 } - \dfrac{ \sqrt{7} + 1}{ \sqrt{7} - 1 } = a + b \sqrt{7} }}$

${ \sf{Taking \: LHS}}$

${ \sf{Rationalise \: \: the \: \: denominator}}$

$= { \sf{ \dfrac{( \sqrt{7 } - 1)( \sqrt{7} - 1)}{( \sqrt{7} +1)( \sqrt{7} - 1)} - \dfrac{( \sqrt{7} + 1 )( \sqrt{7} + 1)}{( \sqrt{7} - 1)( \sqrt{7} + 1)} }}$

${ \sf{Using}} \\ { \sf{\implies {(a + b)}^{2} = {a}^{2} + 2ab + {b}^{2} }} \\ { \sf{ \implies(a + b)(a - b) = {a}^{2} - {b}^{2} }} \\ { \sf{ \implies {(a - b)}^{2} = {a}^{2} - 2ab + {b}^{2} }}$

${ \sf{ = \dfrac{ {( \sqrt{7 } - 1) }^{2} }{ {( \sqrt{7} )}^{2} - {1}^{2} } - \dfrac{ ({ \sqrt{7} + 1) }^{2} }{ {( \sqrt{7} )}^{2} - {1}^{2} } }}$

${ \sf{ = \dfrac{ {( \sqrt{7} - 1) }^{2} - {( \sqrt{7} + 1)}^{2} }{6} }}$

${ \sf{using \: algebraic \:identitiy }} \\ { \sf{\implies {(a + b)}^{2} - {(a - b)}^{2} = 4ab}}$

${ \sf{ = \dfrac{ 4 \sqrt{7} }{6} }}$

${ \sf{a + b \sqrt{7} = \dfrac{2 \sqrt{7} }{3} }}$

${ \sf{a + b \sqrt{7} = 0 + { \dfrac{ 2}{3} \sqrt{7} }^{}}}$

${ \sf{Now \: by \: comparing}}$

${ \sf{a = 0,b = \dfrac{ 2}{3} }}$

Answer 2

Step-by-step explanation:

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