Rationalise the denominator of the following: | vedclass

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(C) To rationalise the denominator,multiply the numerator and the denominator by the conjugate of the denominator,which is $(\sqrt{41}+5)$.$\frac{16}{

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$\sqrt{41}+5$

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(C) To rationalise the denominator,multiply the numerator and the denominator by the conjugate of the denominator,which is $(\sqrt{41}+5)$.$\frac{16}{\sqrt{41}-5} = \frac{16}{\sqrt{41}-5} 	imes \frac{\sqrt{41}+5}{\sqrt{41}+5}$Using the identity $(a-b)(a+b) = a^2 - b^2$ in the denominator:$= \frac{16(\sqrt{41}+5)}{(\sqrt{41})^2 - (5)^2}$$= \frac{16(\sqrt{41}+5)}{41 - 25}$$= \frac{16(\sqrt{41}+5)}{16}$$= \sqrt{41}+5$

Rationalise the denominator of the following:
$\frac{16}{\sqrt{41}-5}$

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Rationalise the denominator of the following:$\frac{16}{\sqrt{41}-5}$