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(D) To rationalise the denominator,multiply the numerator and the denominator by the conjugate of the denominator,which is $(2+\sqrt{3})$.$\frac{2+\sq

Vedclass · Accepted Answer

$7+4\sqrt{3}$

Vedclass · Answer

(D) To rationalise the denominator,multiply the numerator and the denominator by the conjugate of the denominator,which is $(2+\sqrt{3})$.$\frac{2+\sqrt{3}}{2-\sqrt{3}} = \frac{2+\sqrt{3}}{2-\sqrt{3}} 	imes \frac{2+\sqrt{3}}{2+\sqrt{3}}$Using the identity $(a-b)(a+b) = a^2 - b^2$ in the denominator and $(a+b)^2 = a^2 + b^2 + 2ab$ in the numerator:$= \frac{(2+\sqrt{3})^2}{2^2 - (\sqrt{3})^2}$$= \frac{2^2 + (\sqrt{3})^2 + 2(2)(\sqrt{3})}{4 - 3}$$= \frac{4 + 3 + 4\sqrt{3}}{1}$$= 7 + 4\sqrt{3}$

Rationalise the denominator of the following:
$\frac{2+\sqrt{3}}{2-\sqrt{3}}$

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Rationalise the denominator of the following:$\frac{2+\sqrt{3}}{2-\sqrt{3}}$