Question

Rationalise the denominator of
1/4-√3​

Answer 1

Answer:

$\frac{1}{4 - \sqrt{3} } \times \frac{4 + \sqrt{3} }{4 + \sqrt{3} } \\ = > \: \frac{4 + \sqrt{3} }{ {4}^{2} - {( \sqrt{3}) }^{2} } \\ = > \: \frac{4 + \sqrt{3} }{16 - 3} = \frac{4 + \sqrt{3} }{13} \\ \\ hope \: it \: helps \: you....$

Answer 2

$\huge{\underline{\red{\mathfrak{AnSwEr :-}}}}$

$\frac{1}{4 - \sqrt{3} } \\ = \frac{1}{4 - \sqrt{3} } \times \frac{4 + \sqrt{3} }{4 + \sqrt{3} } \\ = \frac{4 + \sqrt{3} }{ ({4})^{2} - ( { \sqrt{3} })^{2} } \\ = \frac{4 + \sqrt{3} }{16 - 3} \\ = \frac{4 + \sqrt{3} }{13}$