Question

4/√7-3 plz gave the answer

Answer 1

Answer:

$by \: rationalisation \: of \: denominator..... \\ \\ = > \frac{4}{ \sqrt{7} - 3 } \\ \\ = > \frac{4 \times ( \sqrt{7} + 3 )}{ (\sqrt{7} - 3) ( \sqrt{7} + 3) } .......(a - b )(a + b) = {a}^{2} - {b}^{2} \\ \\ = > \frac{4 \sqrt{7} + 12}{ {( \sqrt{7}) }^{2} - ({3})^{2} } \\ \\ = > \frac{4 \sqrt{7} + 12}{7 - 9} \\ \\ = > \frac{4 \sqrt{7} + 12 }{ - 2} \\ \\ = > \frac{ - 4 \sqrt{7} + 12 }{2}$

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