\(\frac{5\sqrt{7}-7\sqrt{5}+2\sqrt{70}}{\sqrt{35}}\)

\(=\frac{\sqrt{35}.(5\sqrt{7}-7\sqrt{5}+2\sqrt{70})}{\sqrt{35}.\sqrt{35}}\)

\(=\frac{\sqrt{35}.(5\sqrt{7}-7\sqrt{5}+2\sqrt{70})}{35}\)

\(\sqrt{\frac{4}{3}}+\sqrt{12}-\frac{4}{3}\sqrt{\frac{3}{4}}\)

\(=\frac{\sqrt{4}}{\sqrt{3}}+\sqrt{12}-\frac{4}{3}\cdot\frac{\sqrt{3}}{\sqrt{4}}\)

\(=\frac{2\sqrt{3}}{\sqrt{3}.\sqrt{3}}+\sqrt{12}-\frac{4}{3}\cdot\frac{\sqrt{3}}{2}\)

\(=\frac{2\sqrt{3}}{3}+2\sqrt{3}-\frac{2\sqrt{3}}{3}\)

\(=2\sqrt{3}\left(\frac{1}{3}+1-\frac{1}{3}\right)\)

\(=2\sqrt{3}\)

\(\left(5\sqrt{\frac{1}{5}}+\frac{1}{2}\sqrt{20}-\frac{5}{4}\sqrt{\frac{4}{5}+5}\right):2\sqrt{5}\)

\(=\left(5\cdot\frac{\sqrt{1}}{\sqrt{5}}+\frac{1}{2}\sqrt{4.5}-\frac{5}{4}\sqrt{\frac{4+25}{5}}\right)\cdot\frac{1}{2\sqrt{5}}\)

\(=\left(\frac{5\sqrt{5}}{\sqrt{5}.\sqrt{5}}+\frac{1}{2}.2\sqrt{5}-\frac{5}{4}\sqrt{\frac{29}{5}}\right)\cdot\frac{\sqrt{5}}{2\cdot\sqrt{5}\cdot\sqrt{5}}\)

\(=\left(\frac{5\sqrt{5}}{5}+\sqrt{5}-\frac{5}{4}\cdot\frac{\sqrt{29}}{\sqrt{5}}\right)\cdot\frac{\sqrt{5}}{10}\)

\(=\left(\sqrt{5}+\sqrt{5}-\frac{5}{4}\cdot\frac{\sqrt{29}\sqrt{5}}{\sqrt{5}\sqrt{5}}\right)\cdot\frac{\sqrt{5}}{10}\)

\(=\left(2\sqrt{5}-\frac{5}{4}\cdot\frac{\sqrt{145}}{5}\right)\cdot\frac{\sqrt{5}}{10}\)

\(=\left(2\sqrt{5}-\frac{\sqrt{145}}{4}\right)\cdot\frac{\sqrt{5}}{10}\)

a) \(=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\frac{14}{49-48}=14\)

b) \(=\frac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}-\frac{5\sqrt{6}}{5}+\frac{4\sqrt{3}-12\sqrt{2}}{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}\)

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, = \(=\frac{\sqrt{7}-5}{2}-\frac{3-\sqrt{7}}{2}+\frac{6\sqrt{7}+12}{7-4}-\frac{20-5\sqrt{7}}{16-7}=\frac{\sqrt{7}-5-3+\sqrt{7}}{2}+\frac{6\sqrt{7}+12}{3}-\frac{20-5\sqrt{7}}{9}\)

b. = \(\frac{\sqrt{3}+\sqrt{2}+\sqrt{5}}{\left(\sqrt{3}+\sqrt{2}+\sqrt{5}\right)\left(\sqrt{3}+\sqrt{2}-\sqrt{5}\right)}-\frac{\sqrt{3}+\sqrt{2}-\sqrt{5}}{\left(\sqrt{3}+\sqrt{2}+\sqrt{5}\right)\left(\sqrt{3}+\sqrt{2}-\sqrt{5}\right)}=\frac{\sqrt{3}+\sqrt{2}+\sqrt{5}}{\left(\sqrt{3}+\sqrt{2}\right)^2-\left(\sqrt{5}\right)^2}-\frac{\sqrt{3}+\sqrt{2}-\sqrt{5}}{\left(\sqrt{3}+\sqrt{2}\right)^2-\left(\sqrt{5}\right)^2}\)

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) \(\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\)

\(=\frac{\left(\sqrt{7}+\sqrt{5}\right)^2}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}-\frac{\left(\sqrt{7}-\sqrt{5}\right)^2}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}\)

\(=\frac{\left(\sqrt{7}+\sqrt{5}\right)^2-\left(\sqrt{7}-\sqrt{5}\right)^2}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}\)

\(=\frac{4\sqrt{35}}{2}\)

\(=2\sqrt{35}\)

Bạn xem hộ mk đề cậu b nhé căn 5- căn 2 hay là căn 5 - 2

\(\frac{2}{\sqrt{7}-5}-\frac{2}{\sqrt{7}+5}=\frac{2\sqrt{7}+10}{\left(\sqrt{7}-5\right)\left(\sqrt{7}+5\right)}-\frac{2\sqrt{7}-10}{\left(\sqrt{7}-5\right)\left(\sqrt{7}+5\right)}=\frac{20}{7-25}=\frac{20}{-18}=\frac{10}{-9}\)

\(\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}=\frac{\left(\sqrt{7}+\sqrt{5}\right)^2+\left(\sqrt{7}-\sqrt{5}\right)^2}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}=\frac{12+2\sqrt{35}+12-2\sqrt{35}}{2}=\frac{24}{2}=12\)

\(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right)\frac{1}{\sqrt{7}-\sqrt{5}}=\left(\frac{-\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\frac{-\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right)\frac{1}{\sqrt{7}-\sqrt{5}}=\frac{\left(\sqrt{7}+\sqrt{5}\right)}{\sqrt{5}-\sqrt{7}}=\frac{\left(\sqrt{7}+\sqrt{5}\right)^2}{\left(\sqrt{5}-\sqrt{7}\right)\left(\sqrt{5}+\sqrt{7}\right)}=\frac{12+2\sqrt{35}}{-2}=-6-\sqrt{35}\)

\(\frac{3}{\sqrt{5}-2}+\frac{2}{\sqrt{5}+3}-\frac{1}{\sqrt{5}+4}=\frac{3\left(\sqrt{5}+2\right)}{5-4}+\frac{2\left(\sqrt{5}-3\right)}{5-9}-\frac{\sqrt{5}-4}{5-16}\)

\(=3\sqrt{5}+6+\frac{2\sqrt{5}-6}{-4}+\frac{4-\sqrt{5}}{-11}=\frac{66\sqrt{5}+132}{22}+\frac{33-11\sqrt{5}}{22}+\frac{2\sqrt{5}-8}{22}\)

\(=\frac{66\sqrt{5}-11\sqrt{5}+2\sqrt{5}+132+33-8}{22}=\frac{57\sqrt{5}+157}{22}\)