p: \(F=\dfrac{1}{3}\left(\dfrac{3}{3\cdot6}+\dfrac{3}{6\cdot9}+\dfrac{3}{9\cdot12}+...+\dfrac{3}{30\cdot33}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{10}{33}=\dfrac{10}{99}\)

n: \(F=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)

\(=2\cdot\dfrac{502}{1005}=\dfrac{1004}{1005}\)

m: \(=\left(3-\dfrac{7}{3}+\dfrac{1}{4}\right):\left(4-\dfrac{31}{6}+\dfrac{9}{4}\right)\)

\(=\dfrac{36-28+3}{12}:\dfrac{48-62+27}{12}\)

\(=\dfrac{11}{13}\)

Ta có : \(\dfrac{3\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4.\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}=\dfrac{3}{4}\)

\(\dfrac{\dfrac{3}{41}-\dfrac{12}{47}+\dfrac{27}{53}}{\dfrac{4}{41}-\dfrac{16}{47}+\dfrac{36}{53}}\)

\(=\dfrac{3\times\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4\times\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}\)

\(=\dfrac{3}{4}\)

\(\dfrac{\dfrac{3}{41}-\dfrac{12}{47}+\dfrac{27}{53}}{\dfrac{4}{41}-\dfrac{16}{47}+\dfrac{36}{53}}=\dfrac{3\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}=\dfrac{3}{4}\)

tính nhanh

Ta có: \(F=\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2008\cdot2010}\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)

\(=2\cdot\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\)

\(=2\cdot\dfrac{502}{1005}=\dfrac{1004}{1005}\)

\(F=\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+...+\dfrac{4}{2008.2010}\)

\(F=2.\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{2008.2010}\right)\)

\(F=2.\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)

\(F=2.\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\)

\(F=1-\dfrac{1}{1005}=\dfrac{1004}{1005}\)

=2(2/2.4+2/4.6+......+22/008.2010)

=2(12−12010)

a: \(=\dfrac{13\left(3-18\right)}{40\left(15-2\right)}=\dfrac{13}{15-2}\cdot\dfrac{-15}{40}=\dfrac{-3}{8}\)

b: \(=\dfrac{18\left(34-124\right)}{36\left(-17-13\right)}=\dfrac{1}{2}\cdot\dfrac{-90}{-30}=\dfrac{3}{2}\)

c: \(=\dfrac{3\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}+\dfrac{\dfrac{-1}{4}\cdot\dfrac{-2}{3}-\dfrac{3}{4}:\dfrac{1}{6}}{\dfrac{3}{2}\cdot\left(\dfrac{-2}{3}-\dfrac{3}{4}\cdot\dfrac{-2}{3}\right)}\)

\(=\dfrac{3}{4}+\dfrac{\dfrac{2}{12}-\dfrac{9}{2}}{\dfrac{3}{2}\cdot\dfrac{-1}{6}}=\dfrac{3}{4}+\dfrac{-13}{3}:\dfrac{-3}{12}=\dfrac{3}{4}+\dfrac{13}{3}\cdot\dfrac{12}{3}\)

\(=\dfrac{3}{4}+\dfrac{156}{9}=\dfrac{217}{12}\)

Ta có :

\(F=\dfrac{4}{2.4}+\dfrac{4}{4.6}+..................+\dfrac{4}{2008.2010}\)

\(\Rightarrow F=2\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+.............+\dfrac{2}{2008.2010}\right)\)

\(\Rightarrow F=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+..............+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)

\(\Rightarrow F=2\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\)

\(\Rightarrow F=2.\dfrac{502}{1005}=\dfrac{1004}{1005}\)

\(F=\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+......+\dfrac{4}{2008.2010}\)

\(F=\dfrac{4}{2}\left(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+.....+\dfrac{1}{2008.2010}\right)\)

\(F=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+.....+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)\(F=2\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\)\(F=2.\dfrac{502}{1005}\)

\(F=\dfrac{1004}{1005}\)

\(A=\dfrac{636363\cdot37-373737\cdot63}{1+2+3+...+2006}\)

\(=\dfrac{37^2\cdot3^3\cdot7^2\cdot13-37^2\cdot3^3\cdot7^2\cdot13}{\left(2006+1\right)\cdot1003}\)

=0

\(A=2\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{48.50}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{48}-\dfrac{1}{50}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{50}\right)\)

\(=2\times\dfrac{12}{25}=\dfrac{24}{25}\)

\(=>A=4.\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{46}-\dfrac{1}{48}+\dfrac{1}{48}-\dfrac{1}{50}\right)\)

\(A=4.\left(\dfrac{1}{2}-\dfrac{1}{50}\right)=4.\left(\dfrac{25}{50}-\dfrac{1}{50}\right)=\dfrac{4.24}{50}=\dfrac{48}{25}\)

1: \(\dfrac{11}{24}-\dfrac{5}{41}+\dfrac{13}{24}+0,5-\dfrac{36}{41}\)

\(=\left(\dfrac{11}{24}+\dfrac{13}{24}\right)-\left(\dfrac{5}{41}+\dfrac{36}{41}\right)+\dfrac{1}{2}\)

\(=1-1+\dfrac{1}{2}=\dfrac{1}{2}\)

2: \(12:\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)

\(=12:\left(\dfrac{9}{12}-\dfrac{10}{12}\right)^2\)

\(=12:\left(-\dfrac{1}{12}\right)^2=12:\dfrac{1}{144}=12\cdot144=1368\)

3: \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right)\cdot\left(0,8-\dfrac{3}{4}\right)^2\)

\(=\dfrac{12+8-3}{12}\cdot\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\)

\(=\dfrac{17}{12}\cdot\left(\dfrac{16-15}{20}\right)^2\)

\(=\dfrac{17}{12}\cdot\dfrac{1}{400}=\dfrac{17}{4800}\)

4: \(16\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)+28\dfrac{2}{7}:\dfrac{3}{5}\)

\(=\dfrac{5}{3}\cdot\left(-16-\dfrac{2}{7}\right)+\dfrac{5}{3}\cdot\left(28+\dfrac{2}{7}\right)\)

\(=\dfrac{5}{3}\left(-16-\dfrac{2}{7}+28+\dfrac{2}{7}\right)\)

\(=12\cdot\dfrac{5}{3}=20\)

5: \(\left(2^2:\dfrac{4}{3}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)

\(=\left(4\cdot\dfrac{3}{4}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)

\(=\dfrac{5}{2}\cdot\dfrac{6}{5}-17=3-17=-14\)

6: \(\left(\dfrac{1}{3}\right)^{50}\cdot\left(-9\right)^{25}-\dfrac{2}{3}:4\)

\(=\left(\dfrac{1}{3}\right)^{50}\cdot\left(-1\right)\cdot3^{50}-\dfrac{2}{3\cdot4}\)

\(=-1-\dfrac{2}{12}=-1-\dfrac{1}{6}=-\dfrac{7}{6}\)