Question

1/root7+2+1/root5+root2+2/root6+root5

Answer 1

Simplify:

Step-by-step explanation:

$\ Given \ values: \\\\\frac{1}{\sqrt{7}+2}+\frac{1}{\sqrt{5}+\sqrt{2}}+\frac{2}{\sqrt{6}+\sqrt{5}} \\\\\ Solution:\\\\ \ first \ value: \\ \frac{1}{\sqrt{7}+2} \\\\\frac{1}{\sqrt{7}+2} \times \frac{\sqrt{7}-2}{\sqrt{7}-2}\\\\\frac{\sqrt{7}-2}{(\sqrt{7})^2-(2)^2} \\\\\frac{\sqrt{7}-2}{(7-4)} \\\\ \rightarrow \frac{\sqrt{7}-2}{3}$

$\ second \ value: \\ \frac{1}{\sqrt{5}-\sqrt{2}} \\\\\frac{1}{\sqrt{5}+\sqrt{2}} \times\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}} \\\\\frac{\sqrt{5}-\sqrt{2}}{(\sqrt{5})^2-(\sqrt{2})^2} \\\\\frac{\sqrt{5}-\sqrt{2}}{( 5- 2} \\\\\rightarrow \frac{\sqrt{5}-\sqrt{2}}{3}$

$\ third \ value: \\\frac{2}{\sqrt{6}+\sqrt{5}} \\\\\frac{2}{\sqrt{6}+\sqrt{5}} \times \frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}-\sqrt{5}} \\\\\frac{2(\sqrt{6}-\sqrt{5})}{(\sqrt{6})^2-(\sqrt{5})^2} \\\\\frac{2\sqrt{6}-2\sqrt{5}}{6-5}\\\\ \rightarrow \frac{2\sqrt{6}-2\sqrt{5}}{1}$

$\ add \ values: \\\\\frac{\sqrt{7}-2}{3} +\frac{\sqrt{5}-\sqrt{2}}{3}+\frac{2\sqrt{6}-2\sqrt{5}}{1}\\ \\\frac{\sqrt{7}-2+\sqrt{5}-\sqrt{2}+3(2\sqrt{6}-2\sqrt{5})}{3}\\\\\frac{\sqrt{7}-2+\sqrt{5}-\sqrt{2}+6\sqrt{6}-6\sqrt{5})}{3}\\ \\ \frac{\sqrt{7}-2-5\sqrt{5}-\sqrt{2}+6\sqrt{6}}{3}$

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