Question

rationalise the denominator of each of the following

if you know plz give the answer fast ​

Answer 1

Answer 1:

Given fraction,

$\tt = \dfrac{6}{3 + \sqrt{5} }$

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.

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