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Solution:

(i) \frac{3 + \sqrt{2}}{3 - \sqrt{2}} = a + b \sqrt{2} \Rightarrow Rationalising LHS {Multiply 3 + \sqrt{2} in the numerator and denominator} \Rightarrow \frac{3 + \sqrt{2}}{3 - \sqrt{2}} \times \frac{3 + \sqrt{2}}{3 + \sqrt{2}} \Rightarrow \frac{{(3 + \sqrt{2})}^{2}}{3^{2} - (\sqrt{2)^{2}}} {(a + b) (a - b) = a^{2} - b^{2}} \Rightarrow \frac{9 + 2 + 6 \sqrt{2}}{1} = 11 + 6 \sqrt{2} So, a = 11 and b = 6 (ii) \frac{5 + 2 \sqrt{3}}{7 + 4 \sqrt{3}} = a + b \sqrt{3} \Rightarrow Rationalising LHS {Multiply 7 - 4 \sqrt{3} in the numerator and denominator} \Rightarrow \frac{5 + 2 \sqrt{3}}{7 + 4 \sqrt{3}} \times \frac{7 - 4 \sqrt{3}}{7 - 4 \sqrt{3}} \Rightarrow \frac{35 - 20 \sqrt{3} + 14 \sqrt{3} - 8 \times 3}{7^{2} - {(4 \sqrt{3})}^{2}} \Rightarrow \frac{11 - 6 \sqrt{3}}{49 - 48} = 11 - 6 \sqrt{3} So, a = 11 and b = - 6 (iii) \frac{\sqrt{7} - 1}{\sqrt{7} + 1} - \frac{\sqrt{7} + 1}{\sqrt{7} - 1} = a + b \sqrt{7} \Rightarrow \frac{{(\sqrt{7} - 1)}^{2} - {(\sqrt{7} + 1)}^{2}}{(\sqrt{7} + 1) (\sqrt{7} - 1)} \Rightarrow \frac{- 4 \times \sqrt{7} \times 1}{{(\sqrt{7})}^{2} - 1^{2}} {{(a - b)}^{2} - {(a + b)}^{2} = - 4 ab and (a - b) (a + b) = a^{2} - b^{2}} \Rightarrow \frac{- 4 \sqrt{7}}{7 - 1} = \frac{- 4 \sqrt{7}}{6} = \frac{- 2 \sqrt{7}}{3} So, a = 0 and b = \frac{- 2}{3}

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