Question

Rationalise the denominator of 1/ 7+4√3

Answer 1

$\frac{1}{(7+4\sqrt{3})}\\= \frac{(7-4\sqrt{3})}{(7+4\sqrt{3})(7-4\sqrt{3})}\\ = \frac{(7-4\sqrt{3})}{7^{2}-(4\sqrt{3})^{2}}\\ = \frac{(7-4\sqrt{3})}{49- 48}\\ = 7 - 4\sqrt{3}$

Therefore.,

$\red{ \frac{1}{(7+4\sqrt{3})}}\green{ =7 - 4\sqrt{3}}$

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Answer 2

Answer:-

$= \frac{1}{7 + 4 \sqrt{3} } \\ \\ rationalising \: \: the \: \: denominator..... \\ \\ = \frac{1}{7 + 4 \sqrt{3} } \times \frac{7 - 4 \sqrt{3} }{7 - 4 \sqrt{3} } \\ \\ = \frac{7 - 4 \sqrt{3} }{ {(7)}^{2} - {(4 \sqrt{3} })^{2} } \\ \\ = \frac{7 - 4 \sqrt{3} }{49 - 48} \\ \\ = 7 + - 4 \sqrt{3} ...$