Question

root2+ root3/ root5+root6 rationalized
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Answer 1

Answer:

$2\sqrt{3}+3\sqrt{2}-\sqrt{10}-\sqrt{15}$

Step-by-step explanation:

$\frac{\sqrt{2} +\sqrt{3}} {\sqrt{5} +\sqrt{6}}$

In order to rationalize it, we multiply the denominator by its conjugate

This way we have a difference of squares

$\boxed{(a+b)(a-b) = a^2-b^2}$

$\frac{\sqrt{2} +\sqrt{3}} {\sqrt{5} +\sqrt{6}} \cdot \frac{\sqrt{5} -\sqrt{6}} {\sqrt{5} -\sqrt{6}} = \frac{(\sqrt{2} +\sqrt{3})(\sqrt{5} -\sqrt{6})}{5-6}$

$\frac{(\sqrt{2} +\sqrt{3})(\sqrt{5} -\sqrt{6})}{5-6}=\frac{\left(\sqrt{10}-2\sqrt{3}+\sqrt{15}-3\sqrt{2}\right)}{-1}$

$-(\sqrt{10}-2\sqrt{3}+\sqrt{15}-3\sqrt{2}) = 2\sqrt{3}+3\sqrt{2}-\sqrt{10}-\sqrt{15}$