Rationalise the denominator of frac{1}{7+3 sq | vedclass

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Question

(A) To rationalise the denominator of $\frac{1}{7+3 \sqrt{2}}$,we multiply the numerator and the denominator by the conjugate of the denominator,which

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$\frac{7-3 \sqrt{2}}{31}$

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(A) To rationalise the denominator of $\frac{1}{7+3 \sqrt{2}}$,we multiply the numerator and the denominator by the conjugate of the denominator,which is $(7-3 \sqrt{2})$.$\frac{1}{7+3 \sqrt{2}} = \frac{1}{7+3 \sqrt{2}} 	imes \frac{7-3 \sqrt{2}}{7-3 \sqrt{2}}$Using the identity $(a+b)(a-b) = a^2 - b^2$ in the denominator:$= \frac{7-3 \sqrt{2}}{(7)^2 - (3 \sqrt{2})^2}$$= \frac{7-3 \sqrt{2}}{49 - (9 	imes 2)}$$= \frac{7-3 \sqrt{2}}{49 - 18}$$= \frac{7-3 \sqrt{2}}{31}$

Rationalise the denominator of $\frac{1}{7+3 \sqrt{2}}$

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