Simplify: frac{1}{sqrt{6}-sqrt{5}}-frac{3}{sq | vedclass

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(A) To simplify the expression,we rationalize each term individually:$1$. $\frac{1}{\sqrt{6}-\sqrt{5}} = \frac{1(\sqrt{6}+\sqrt{5})}{(\sqrt{6}-\sqrt{5

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(A) To simplify the expression,we rationalize each term individually:$1$. $\frac{1}{\sqrt{6}-\sqrt{5}} = \frac{1(\sqrt{6}+\sqrt{5})}{(\sqrt{6}-\sqrt{5})(\sqrt{6}+\sqrt{5})} = \frac{\sqrt{6}+\sqrt{5}}{6-5} = \sqrt{6}+\sqrt{5}$$2$. $\frac{3}{\sqrt{5}-\sqrt{2}} = \frac{3(\sqrt{5}+\sqrt{2})}{(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})} = \frac{3(\sqrt{5}+\sqrt{2})}{5-2} = \frac{3(\sqrt{5}+\sqrt{2})}{3} = \sqrt{5}+\sqrt{2}$$3$. $\frac{4}{\sqrt{6}+\sqrt{2}} = \frac{4(\sqrt{6}-\sqrt{2})}{(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})} = \frac{4(\sqrt{6}-\sqrt{2})}{6-2} = \frac{4(\sqrt{6}-\sqrt{2})}{4} = \sqrt{6}-\sqrt{2}$Now,substitute these back into the original expression:$(\sqrt{6}+\sqrt{5}) - (\sqrt{5}+\sqrt{2}) - (\sqrt{6}-\sqrt{2})$$= \sqrt{6} + \sqrt{5} - \sqrt{5} - \sqrt{2} - \sqrt{6} + \sqrt{2}$$= (\sqrt{6} - \sqrt{6}) + (\sqrt{5} - \sqrt{5}) + (\sqrt{2} - \sqrt{2})$$= 0 + 0 + 0 = 0$

Simplify: $\frac{1}{\sqrt{6}-\sqrt{5}}-\frac{3}{\sqrt{5}-\sqrt{2}}-\frac{4}{\sqrt{6}+\sqrt{2}}$

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