Rationalise the denominator of each of the followingif you know plz gi

1.	Rationalise the denominator of each of the followingif you know plz give the answer fast
Answer» ANSWER 1: Given FRACTION, $\tt = \dfrac{6}{<klux>3</klux> + \sqrt{5} }$ Multiplying both numerator and denominator by (3 - √5), we GET, $\tt = \dfrac{6(3 - \sqrt{5}) }{(3 + \sqrt{5})(3 - \sqrt{5})}$ Using identity a² - b² = (a + b)(a - b), we get, $\tt = \dfrac{6(3 - \sqrt{5}) }{(3)^{2} - (\sqrt{5})^{2}}$ $\tt = \dfrac{6(3 - \sqrt{5}) }{9- 5}$ $\tt = \dfrac{6(3 - \sqrt{5}) }{4}$ $\tt = \dfrac{3(3 - \sqrt{5}) }{2}$ Which is our required answer. Answer 2: Given fraction, $\tt = \dfrac{2}{ \sqrt{3} - 1}$ Multiplying both numerator and denominator by (√3 + 1), we get, $\tt = \dfrac{2( \sqrt{3} + 1) }{(\sqrt{3} - 1)( \sqrt{3} + 1) }$ Using identity (a + b)(a - b) = a² - b², we get, $\tt = \dfrac{2( \sqrt{3} + 1) }{(\sqrt{3})^{2} - (1)^{2}}$ $\tt = \dfrac{2( \sqrt{3} + 1) }{3 - 1}$ $\tt = \dfrac{2( \sqrt{3} + 1) }{2}$ $\tt = \sqrt{3} + 1$ Which is our required answer. Answer 3: Given fraction, $\tt = \dfrac{2}{ \sqrt{3} + \sqrt{2} }$ Multiplying both numerator and denominator by (√3 - √2), we get, $\tt = \dfrac{2( \sqrt{3} - \sqrt{2} ) }{( \sqrt{3} + \sqrt{2} )( \sqrt{3} - \sqrt{2})}$ Using identity (a + b)(a - b) = a² - b², we get, $\tt = \dfrac{2( \sqrt{3} - \sqrt{2} ) }{( \sqrt{3})^{2} -(\sqrt{2} )^{2} }$ $\tt = \dfrac{2( \sqrt{3} - \sqrt{2} ) }{3 - 2}$ $\tt = \dfrac{2( \sqrt{3} - \sqrt{2} ) }{1}$ $\tt = 2( \sqrt{3} - \sqrt{2})$ Which is our required answer. •••♪