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\(\sqrt{1+\frac{8n^2-1}{\left(2n-1\right)^2\left(2n+1\right)^2}}=\sqrt{1+\frac{8n^2-1}{\left(4n^2-1\right)^2}}=\sqrt{\frac{\left(4n^2-1\right)^2+8n^2-1}{\left(4n^2-1\right)^2}}\)
\(=\sqrt{\frac{16n^4-8n^2+1+8n^2-1}{\left(4n^2-1\right)^2}}=\frac{4n^2}{4n^2-1}=1+\frac{1}{4n^2-1}=1+\frac{1}{2}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)\)
\(\Rightarrow S=1009+\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2017}-\frac{1}{2019}\right)\)
\(=1009+\frac{1}{2}\left(1-\frac{1}{2019}\right)=...\)
a)\(\left(\sqrt{12}+\sqrt{75}+\sqrt{27}\right):\sqrt{15}\)
\(=\left(2\sqrt{3}+5\sqrt{3}+3\sqrt{3}\right):\sqrt{15}\)
\(=10\sqrt{3}:\sqrt{15}=\sqrt{300}:\sqrt{15}=\sqrt{20}=2\sqrt{5}\)
b) \(\frac{12\sqrt{50}-8\sqrt{200}+7\sqrt{450}}{\sqrt{10}}\)
\(=\frac{60\sqrt{2}-80\sqrt{2}+105\sqrt{2}}{\sqrt{10}}\)
\(=\frac{85\sqrt{2}}{10}=\frac{17\sqrt{2}}{2}\)
c)\(\frac{\sqrt{\frac{1}{7}}-\sqrt{\frac{16}{7}}+\sqrt{\frac{9}{7}}}{7}=\frac{\frac{1}{\sqrt{7}}-\frac{4}{\sqrt{7}}+\frac{3}{\sqrt{7}}}{7}=\frac{0}{7}=0\)
a, = \(\frac{\sqrt{7}-5}{2}-\frac{2\left(3-\sqrt{7}\right)}{4}+\frac{6\left(\sqrt{7}+2\right)}{\left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right)}-\frac{5\left(4-\sqrt{7}\right)}{\left(4-\sqrt{7}\right)\left(4+\sqrt{7}\right)}\)
a) \(\frac{2}{4-3\sqrt{2}}-\frac{2}{4+3\sqrt{2}}\)
\(=\frac{2\left(4+3\sqrt{2}\right)}{\left(4-3\sqrt{2}\right)\left(4+3\sqrt{2}\right)}-\frac{2\left(4-3\sqrt{2}\right)}{\left(4-3\sqrt{2}\right)\left(4+3\sqrt{2}\right)}\)
\(=\frac{2\left(4+3\sqrt{2}\right)-2\left(4-3\sqrt{2}\right)}{\left(4-3\sqrt{2}\right)\left(4+3\sqrt{2}\right)}\)
\(=\frac{12\sqrt{2}}{-2}\)
\(=-6\sqrt{2}\)
b) \(\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\)
\(=\frac{\left(\sqrt{7}+\sqrt{5}\right)^2}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}-\frac{\left(\sqrt{7}-\sqrt{5}\right)^2}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}\)
\(=\frac{\left(\sqrt{7}+\sqrt{5}\right)^2-\left(\sqrt{7}-\sqrt{5}\right)^2}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}\)
\(=\frac{4\sqrt{35}}{2}\)
\(=2\sqrt{35}\)
a) \(\sqrt{200}+2\sqrt{108}-\sqrt{98}+\frac{1}{3}\sqrt{\frac{81}{3}}-3\sqrt{75}\)
\(=10\sqrt{2}+12\sqrt{3}-7\sqrt{2}+\sqrt{3}-15\sqrt{3}\)
\(=3\sqrt{2}-2\sqrt{3}\)
b)\(\left(21\sqrt{\frac{1}{7}}+\frac{1}{2}\sqrt{112}-\frac{14}{3}\sqrt{\frac{9}{7}}+7\right):3\sqrt{7}\)
\(=\left(3\sqrt{7}+2\sqrt{7}-2\sqrt{7}+7\right):3\sqrt{7}\)
\(=\frac{\sqrt{7}\left(3+\sqrt{7}\right)}{3\sqrt{7}}=\frac{\sqrt{7}+3}{3}\)
c)\(\left(\sqrt{27}-\sqrt{125}+\sqrt{45}+\sqrt{12}\right)\left(\sqrt{75}+\sqrt{20}\right)\)
\(=\left(3\sqrt{3}-5\sqrt{5}+3\sqrt{5}+2\sqrt{3}\right)\left(5\sqrt{3}+2\sqrt{5}\right)\)
\(=\left(5\sqrt{3}-2\sqrt{5}\right)\left(5\sqrt{3}+2\sqrt{5}\right)\)
\(=75-20=55\)
d)\(\left(\frac{3}{\sqrt{6}-3}-\frac{3}{\sqrt{6}+3}\right).\frac{3-\sqrt{3}}{2-2\sqrt{3}}-\frac{\sqrt{28-6\sqrt{3}}}{1}\)
\(=\frac{3\left(\sqrt{6}+3\right)-3\left(\sqrt{6}-3\right)}{-3}.\frac{3-\sqrt{3}}{2-2\sqrt{3}}-\sqrt{\left(3\sqrt{3}-1\right)^2}\)
\(=\frac{-6\left(3-\sqrt{3}\right)}{2-2\sqrt{3}}-\left(3\sqrt{3}-1\right)\left(do3\sqrt{3}>1\right)\)
\(=\frac{6\sqrt{3}-18}{2-2\sqrt{3}}-\frac{8\sqrt{3}-20}{2-2\sqrt{3}}\)
\(=\frac{6\sqrt{3}-18-8\sqrt{3}+20}{2-2\sqrt{3}}=\frac{2-2\sqrt{3}}{2-2\sqrt{3}}=1\)