IDNLearn.com: Where questions are met with accurate and insightful answers. Find reliable solutions to your questions quickly and accurately with help from our dedicated community of experts.
Sagot :
[tex]\frac{4}{2+\sqrt3+\sqrt7}\cdot\frac{2-(\sqrt3+\sqrt7)}{2-(\sqrt3+\sqrt7)}=\frac{8-4\sqrt3-4\sqrt7}{2^2-(\sqrt3+\sqrt7)^2}=\frac{8-4\sqrt3-4\sqrt7}{4-3-2\sqrt{3\cdot7}-7}\\\\=\frac{8-4\sqrt3-4\sqrt7}{-6-2\sqrt{21}}=\frac{-2(2\sqrt3+2\sqrt7-4)}{-2(3+\sqrt{21})}=\frac{2\sqrt3+2\sqrt7-4}{3+\sqrt{21}}\cdot\frac{3-\sqrt{21}}{3-\sqrt{21}}\\\\=\frac{6\sqrt3-2\sqrt{63}+6\sqrt7-2\sqrt{147}-12+4\sqrt{21}}{3^2-(\sqrt{21})^2}=\frac{6\sqrt3-2\sqrt{9\cdot7}+6\sqrt7-2\sqrt{49\cdot3}-12+4\sqrt{21}}{9-21}[/tex]
[tex]=\frac{6\sqrt3-6\sqrt7+6\sqrt7-14\sqrt3-12+4\sqrt{21}}{-12}=\frac{-8\sqrt3+4\sqrt{21}-12}{-12}=\frac{-4(2\sqrt3-\sqrt{21}+3)}{-12}\\\\=\frac{2\sqrt3-\sqrt{21}+3}{3}[/tex]
==============================================================
[tex]\frac{4}{2\sqrt3+\sqrt5}\cdot\frac{2\sqrt3-\sqrt5}{2\sqrt3-\sqrt5}=\frac{8\sqrt3-4\sqrt5}{(2\sqrt3)^2-(\sqrt5)^2}=\frac{8\sqrt3-4\sqrt5}{4\cdot3-5}=\frac{8\sqrt3-4\sqrt5}{12-5}\\\\=\frac{8\sqrt3-4\sqrt5}{7}[/tex]
[tex]=\frac{6\sqrt3-6\sqrt7+6\sqrt7-14\sqrt3-12+4\sqrt{21}}{-12}=\frac{-8\sqrt3+4\sqrt{21}-12}{-12}=\frac{-4(2\sqrt3-\sqrt{21}+3)}{-12}\\\\=\frac{2\sqrt3-\sqrt{21}+3}{3}[/tex]
==============================================================
[tex]\frac{4}{2\sqrt3+\sqrt5}\cdot\frac{2\sqrt3-\sqrt5}{2\sqrt3-\sqrt5}=\frac{8\sqrt3-4\sqrt5}{(2\sqrt3)^2-(\sqrt5)^2}=\frac{8\sqrt3-4\sqrt5}{4\cdot3-5}=\frac{8\sqrt3-4\sqrt5}{12-5}\\\\=\frac{8\sqrt3-4\sqrt5}{7}[/tex]
[tex](1) \frac{4}{2+\sqrt{3} +\sqrt{7}} \\ \\ or, \frac{4}{2+\sqrt{3} +\sqrt{7}} * \frac{2 - \sqrt{3} -\sqrt{7}}{2-\sqrt{3}-\sqrt{7}} \\ \\ => \frac{ \sqrt[2]{3} - \sqrt{21}+3}{3} \\ \\ \\ (2) \frac{4}{\sqrt[2]{3} + \sqrt{5}} \\ \\ or, \frac{4}{\sqrt[2]{3} + \sqrt{5}} * \frac{\sqrt[2]{3}-\sqrt{5}}{\sqrt[2]{3}-\sqrt{5}} \\ \\ => \frac{\sqrt[8]{3}-\sqrt[4]{5}}{7} [/tex]
Thank you for using this platform to share and learn. Don't hesitate to keep asking and answering. We value every contribution you make. IDNLearn.com has the solutions to your questions. Thanks for stopping by, and see you next time for more reliable information.
