Simplify: frac{7 sqrt{3}}{sqrt{10}+sqrt{3}}-f | vedclass

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(D) To simplify the expression,we rationalize the denominator for each term:$\frac{7 \sqrt{3}}{\sqrt{10}+\sqrt{3}} 	imes \frac{\sqrt{10}-\sqrt{3}}{\s

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(D) To simplify the expression,we rationalize the denominator for each term:$\frac{7 \sqrt{3}}{\sqrt{10}+\sqrt{3}} 	imes \frac{\sqrt{10}-\sqrt{3}}{\sqrt{10}-\sqrt{3}} = \frac{7 \sqrt{3}(\sqrt{10}-\sqrt{3})}{10-3} = \frac{7 \sqrt{3}(\sqrt{10}-\sqrt{3})}{7} = \sqrt{30}-3$$\frac{2 \sqrt{5}}{\sqrt{6}+\sqrt{5}} 	imes \frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}-\sqrt{5}} = \frac{2 \sqrt{5}(\sqrt{6}-\sqrt{5})}{6-5} = 2 \sqrt{30}-10$$\frac{3 \sqrt{2}}{\sqrt{15}+3 \sqrt{2}} 	imes \frac{\sqrt{15}-3 \sqrt{2}}{\sqrt{15}-3 \sqrt{2}} = \frac{3 \sqrt{2}(\sqrt{15}-3 \sqrt{2})}{15-18} = \frac{3 \sqrt{30}-18}{-3} = -\sqrt{30}+6$Now,substitute these back into the original expression:$(\sqrt{30}-3) - (2 \sqrt{30}-10) - (-\sqrt{30}+6)$$= \sqrt{30}-3-2 \sqrt{30}+10+\sqrt{30}-6$$= (\sqrt{30}-2 \sqrt{30}+\sqrt{30}) + (-3+10-6)$$= 0 + 1 = 1$

Simplify: $\frac{7 \sqrt{3}}{\sqrt{10}+\sqrt{3}}-\frac{2 \sqrt{5}}{\sqrt{6}+\sqrt{5}}-\frac{3 \sqrt{2}}{\sqrt{15}+3 \sqrt{2}}$

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