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Question Number 200169 by hardmath last updated on 15/Nov/23

$$\mathrm{Rationalise}\mathrm{the}\mathrm{deniminator}\mathrm{of}\mathrm{the}$$$$\mathrm{following}\mathrm{fraction}:$$$$\frac{1}{\sqrt{6}-\sqrt{3}+\sqrt{2}+1}=?$$

Answered by Sutrisno last updated on 15/Nov/23

$$\frac{1}{\sqrt{6}-\sqrt{3}+\sqrt{2}+1}\times \frac{(\sqrt{6}-\sqrt{3})-(\sqrt{2}+1)}{(\sqrt{6}-\sqrt{3})-(\sqrt{2}+1)}$$$$\frac{(\sqrt{6}-\sqrt{3})-(\sqrt{2}+1)}{(6-2\sqrt{18}+3)-(2+2\sqrt{2}+1)}$$$$\frac{(\sqrt{6}-\sqrt{3})-(\sqrt{2}+1)}{6-8\sqrt{2}}\times \frac{6+8\sqrt{2}}{6+8\sqrt{2}}$$$$\frac{((\sqrt{6}-\sqrt{3})-(\sqrt{2}+1))(6+8\sqrt{2})}{-92}$$

Commented by Frix last updated on 15/Nov/23

$$=\frac{6\sqrt{6}-11\sqrt{3}+14\sqrt{2}+22}{92}$$

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