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(A) To rationalise the denominator of $\frac{1}{3+\sqrt{8}}$,we multiply the numerator and the denominator by the conjugate of the denominator,which i

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$3-\sqrt{8}$

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(A) To rationalise the denominator of $\frac{1}{3+\sqrt{8}}$,we multiply the numerator and the denominator by the conjugate of the denominator,which is $(3-\sqrt{8})$.$\frac{1}{3+\sqrt{8}} = \frac{1}{3+\sqrt{8}} 	imes \frac{3-\sqrt{8}}{3-\sqrt{8}}$Using the identity $(a+b)(a-b) = a^2 - b^2$ in the denominator:$= \frac{3-\sqrt{8}}{(3)^2 - (\sqrt{8})^2}$$= \frac{3-\sqrt{8}}{9-8}$$= \frac{3-\sqrt{8}}{1}$$= 3-\sqrt{8}$

The value obtained after rationalising the denominator of $\frac{1}{3+\sqrt{8}}$ is $\ldots \ldots$

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