Question

plz solve this rationalise the denominator question​

Answer 1

Given :

Rationalise the denominator of $\sf{\dfrac{\sqrt{5}- \sqrt{3}}{2- \sqrt{3}} }$

Solution :

$\\ \bullet \bf \frac{ \sqrt{5} - \sqrt{3} }{2 - \sqrt{3} }$

• By multiplying in numerator and denominator by $\bf{2+ \sqrt{3}}$

$\\ \implies \sf \: \frac{ \sqrt{5} - \sqrt{3} }{2 - \sqrt{3} }$

$\\ \implies \sf \: \frac{ \sqrt{5} - \sqrt{3} }{2 - \sqrt{3} } \times \frac{2 + \sqrt{3} }{2 + \sqrt{3} }$

$\\ \implies \sf \: \frac{( \sqrt{5} - \sqrt{3} )(2 + \sqrt{3} )}{(2 - \sqrt{3} )(2 + \sqrt{3}) }$

$\\ \implies \sf \: \frac{ \sqrt{5}(2 + \sqrt{3}) - \sqrt{3} (2 + \sqrt{3} )}{ {(2)}^{2} - ( { \sqrt{3} })^{2} }$

$\\ \implies \sf \: \frac{2 \sqrt{5} + \sqrt{15} - \sqrt{6} - 3 }{4 - 3}$

$\implies \boxed{\bf \: 2 \sqrt{5} + \sqrt{15} - \sqrt{6} - 3}$

By rationalizing the numerator and denominator we get,

$\\ \red \bigstar \boxed{ \sf{2 \sqrt{5} + \sqrt{15} - \sqrt{6} - 3}}$

Answer 2

★ Question:-

Rationalise the denominator of $\sf{\dfrac{\sqrt{5}- \sqrt{3}}{2- \sqrt{3}} }$

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━

★ Given :-

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

★ Solution :-

➟ Find The conjugate irrational number of denominator.

• Multiply $\bf{2+ \sqrt{3}}$ in numerator and denominator

$\\ \implies \sf \: \frac{ \sqrt{5} - \sqrt{3} }{2 - \sqrt{3} }$

$\\ \implies \sf \: \frac{ \sqrt{5} - \sqrt{3} }{2 - \sqrt{3} } \times \frac{2 + \sqrt{3} }{2 + \sqrt{3} }$

$\\ \implies \sf \: \frac{( \sqrt{5} - \sqrt{3} )(2 + \sqrt{3} )}{(2 - \sqrt{3} )(2 + \sqrt{3}) }$

$\\ \implies \sf \: \frac{ \sqrt{5}(2 + \sqrt{3}) - \sqrt{3} (2 + \sqrt{3} )}{ {(2)}^{2} - ( { \sqrt{3} })^{2} }$

$\\ \implies \sf \: \frac{2 \sqrt{5} + \sqrt{15} - \sqrt{6} - 3 }{4 - 3}$

$\\ ➡\bf \: 2 \sqrt{5} + \sqrt{15} - \sqrt{6} - 3$

➟ rationalizing the numerator and denominator

★ Required Answer :-

$\\ \pink\bigstar { \sf{2 \sqrt{5} + \sqrt{15} - \sqrt{6} - 3}}$

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