Question
Rationalise the denominators of : $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
✓
Answer
$ \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$
$ =\frac{(\sqrt{3}-\sqrt{2})^2}{(\sqrt{3})^2-(\sqrt{2})^2}$
$ =\frac{3+2-2 \sqrt{6}}{3-2}$
$ =5-2 \sqrt{ } 6$
$ =\frac{(\sqrt{3}-\sqrt{2})^2}{(\sqrt{3})^2-(\sqrt{2})^2}$
$ =\frac{3+2-2 \sqrt{6}}{3-2}$
$ =5-2 \sqrt{ } 6$
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