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Question

Rationalize the denominators of the following: 
i) $$\cfrac { 1 }{ 3+\sqrt { 2 }  } $$
ii) $$\cfrac { 1 }{ \sqrt { 7 } -\sqrt { 6 }  } $$
iii) $$\cfrac { 1 }{ \sqrt { 7 }  } $$


Solution

(i)$$\dfrac{1}{3+\sqrt{2}}=\dfrac{1}{3+\sqrt{2}}\times \dfrac{3-\sqrt{2}}{3-\sqrt{2}}=\dfrac{3-\sqrt{2}}{(3)^{2}-(\sqrt{2})^{2}}=\dfrac{3-\sqrt{2}}{7}$$

(ii)$$\dfrac{1}{\sqrt{7}-\sqrt{6}}=\dfrac{1}{\sqrt{7}-\sqrt{6}}\times \dfrac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}=\dfrac{\sqrt{7}+\sqrt{6}}{(\sqrt{7})^{2}-(\sqrt{6})^{2}}=\sqrt{7}+\sqrt{6}$$

(iii)$$\dfrac{1}{\sqrt{7}}=\dfrac{1}{\sqrt{7}}\times \dfrac{\sqrt{7}}{\sqrt{7}}=\dfrac{\sqrt{7}}{7}$$

Mathematics

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