Question

answer the question above

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Answer 1

$\large\underline\mathfrak\red{Solution \: :- }$

$\sf{ \frac{ \sqrt{5} - 2}{ \sqrt{5} + 2 } - \frac{ \sqrt{5} + 2}{ \sqrt{5} - 2} }$

$\sf{ \frac{( \sqrt{5} - 2) \times (\sqrt{5} - 2) - ( \sqrt{5} + 2) \times (\sqrt{5} + 2)}{( \sqrt{5 } + 2)( \sqrt{5} - 2) } }$

$\sf\underline\green{Using \: 3 \: Formula's \: :- }$

1) (a+b)² = a²+2ab+b²

2) (a-b)² = a²-2ab+b²

3) (a+b) (a-b) = a²-b²

$\sf{ \frac{( \sqrt{5} - 2) {}^{2} - ( \sqrt{5} + 2) {}^{2}}{( \sqrt{5}) {}^{2} - (2) {}^{2} } }$

$\sf{ \frac{( \sqrt{5}) {}^{2} - ( 2) \sqrt{5} (2) + (2) {}^{2} - ( \sqrt{5} + 2) {}^{2} }{( \sqrt{5} ) {}^{2} - (2) {}^{2} } }$

$\sf{ \frac{5 - 4 \sqrt{5} + 4 - (( \sqrt{5 } ) {}^{2} + (2) \sqrt{5} (2) + (2) {}^{2} }{( \sqrt{5} ) {}^{2} - (2) {}^{2} } }$

$\sf{ \frac{9 - 4 \sqrt{5} - (5 + 4 \sqrt{5} + 4)}{( \sqrt{5}) {}^{2} - (2) {}^{2} } }$

$\sf{ \frac{9 - 4 \sqrt{5} + 9 + 4 \sqrt{5} }{5 - 4} }$

$\sf{ \frac{9 + 9}{1} }$

$\sf{\frac{18}{1} = 18}$

Answer = 18

Sorry friend correct answer -8√5

Sorry I given by mistake some correction in answer

How i tell

In numerator

5-4√5+4-(5+4√5+4)

Because of sign - in bracket sign change

5-4√5+4-5-4√5-4

-4√5-4√5

Answer :- -8√5

Answer 2

$\sf \bf \huge {\boxed {\mathbb {QUESTION}}}$

$\tt Simplify\:\:\dfrac{\sqrt{5}-2}{\sqrt{5}+2} - \dfrac{\sqrt{5}+2}{\sqrt{5}-2}$

$\sf \bf \huge {\boxed {\mathbb {SOLUTION}}}$

$\bf \dfrac{\sqrt{5}-2}{\sqrt{5}+2} - \dfrac{\sqrt{5}+2}{\sqrt{5}-2}$

${\underbrace {\overbrace {\orange {\pmb {By\:taking\: LCM}}}}}$

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

$\bf \implies \dfrac{{\big(\sqrt{5}-2\big)}^{2}-{\big(\sqrt{5}+2\big)}^{2}}{{\big(\sqrt{5}\big)}^{2}-{\big(2\big)}^{2}}\longrightarrow \bigg[\because (a+b) (a-b) ={a}^{2}-{b}^{2} \bigg]$

$\bf \implies \dfrac{\Big\{{\big(\sqrt{5}\big)}^{2}-2\big(\sqrt{5}\big)\big(2\big)+{\big(2\big)}^{2}\Big\}-\Big\{{\big(\sqrt{5}\big)}^{2}+2\big(\sqrt{5}\big)\big(2\big)+{\big(2\big)}^{2}\Big\}}{5-4}\longrightarrow \bigg[\because {(a-b)}^{2}={a}^{2}-2ab+{b}^{2}\:and\:{(a+b)}^{2}={a}^{2}+2ab+{b}^{2} \bigg]$

$\bf \implies \dfrac{\Big\{5-4\sqrt{5}+4\Big\}-\Big\{5+4\sqrt{5}+4\Big\}}{1}$

$\bf \implies \Big\{9-4\sqrt{5}\Big\}-\Big\{9+4\sqrt{5}\Big\}$

$\bf \implies 9-4\sqrt{5}-9-4\sqrt{5}$

$\bf \implies \cancel{9}-4\sqrt{5}\cancel{-9}-4\sqrt{5}$

$\bf \implies -4\sqrt{5}-4\sqrt{5}$

$\bf \implies 2\Big(-4\sqrt{5}\Big)$

${\implies {\blue {\boxed {\boxed {\purple {\mathfrak {-8\sqrt{5}}}}}}}}$

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$\sf \bf \huge {\boxed {\mathbb {EXTRA\:INFORMATION}}}$

${\pink {\sf {Identities}}}$

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

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

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