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A = \(\left(-2\right).\left(-1\dfrac{1}{2}\right).\left(-1\dfrac{1}{3}\right).\left(-1\dfrac{1}{4}\right)...\left(-1\dfrac{1}{214}\right)\)
= \(\left(-2\right).\left(-\dfrac{3}{2}\right).\left(-\dfrac{4}{3}\right).\left(-\dfrac{5}{4}\right)...\left(-\dfrac{215}{214}\right)\)
= \(\dfrac{\left(-2\right).\left(-3\right).\left(-4\right).\left(-5\right)...\left(-215\right)}{1.2.3.4...214}\)
= \(\dfrac{2.3.4.5...215}{1.2.3.4...214}\)
= \(\dfrac{215}{1}=215\)
B = \(\left(-1\dfrac{1}{2}\right).\left(-1\dfrac{1}{3}\right).\left(-1\dfrac{1}{4}\right)....\left(-1\dfrac{1}{299}\right)\)
= \(\left(-\dfrac{3}{2}\right).\left(-\dfrac{4}{3}\right).\left(-\dfrac{5}{4}\right)...\left(-\dfrac{300}{299}\right)\)
= \(\dfrac{\left(-3\right).\left(-4\right).\left(-5\right)...\left(-300\right)}{2.3.4...299}\)
= \(\dfrac{3.4.5...300}{2.3.4.5...299}\)
= \(\dfrac{300}{2}=150\)
\(A=\left(-2\right)\left(-1\frac{1}{2}\right).\left(-1\frac{1}{3}\right).\left(-1\frac{1}{4}\right)...\left(-1\frac{1}{214}\right)\)
\(=2.\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{215}{214}=215\)
\(B=\left(-1\frac{1}{2}\right).\left(-1\frac{1}{3}\right).\left(-1\frac{1}{4}\right)....\left(-1\frac{1}{299}\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{300}{299}=\frac{300}{2}=150\)
\(C=-\frac{7}{4}\left(\frac{33}{12}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{333333}{424242}\right)\)
\(=-\frac{7}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(=-\frac{7}{4}.33.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
\(=-\frac{231}{4}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(=-\frac{231}{4}\left(\frac{1}{3}-\frac{1}{7}\right)\)
\(=-\frac{231}{4}.\frac{4}{21}=-11\)
\(B=\dfrac{7}{4}\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{30}+\dfrac{33}{42}\right)\)
\(=\dfrac{7}{4}\left(\dfrac{11}{4}+\dfrac{33}{20}+\dfrac{11}{10}+\dfrac{11}{14}\right)\)
\(=\dfrac{7}{4}\cdot\dfrac{11\cdot35+33\cdot7+11\cdot14+11\cdot10}{140}\)
\(=\dfrac{880}{20\cdot4}=11\)
\(C=\dfrac{\left(\dfrac{53}{4}-\dfrac{59}{27}-\dfrac{65}{6}\right)\cdot\dfrac{5751}{25}+\dfrac{187}{4}}{\dfrac{100}{21}:\dfrac{-41}{21}}\)
\(=\dfrac{\dfrac{25}{108}\cdot\dfrac{5751}{25}+\dfrac{187}{4}}{\dfrac{-100}{41}}\)
\(=\dfrac{\dfrac{5751+187\cdot27}{108}}{\dfrac{-100}{41}}=100\cdot\dfrac{-41}{100}=-41\)
\(D=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{19}-\dfrac{1}{20}=\dfrac{1}{2}-\dfrac{1}{20}=\dfrac{9}{20}\)
\(E=\dfrac{1}{99}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{98\cdot99}\right)\)
\(=\dfrac{1}{99}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{98}-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{99}-1+\dfrac{1}{99}=\dfrac{2}{99}-1=-\dfrac{97}{99}\)
\(b,C=\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\\ =\dfrac{1}{3.6}+\dfrac{1}{6.9}+\dfrac{1}{9.12}+...+\dfrac{1}{30.33}\\ =\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{12}+...+\dfrac{1}{30}-\dfrac{1}{33}\\ =\dfrac{1}{3}-\dfrac{1}{33}\\ =\dfrac{11}{33}-\dfrac{1}{33}=\dfrac{10}{33}\)
a.F=\(\dfrac{4}{2.4}\)+\(\dfrac{4}{4.6}\)+\(\dfrac{4}{6.8}\)+...+\(\dfrac{4}{2008.2010}\)
F=\(\dfrac{2.2}{2.4}\)+\(\dfrac{2.2}{4.6}\)+\(\dfrac{2.2}{6.8}\)+...+\(\dfrac{2.2}{2008.2010}\)
F=2.(\(\dfrac{2}{2.4}\)+\(\dfrac{2}{4.6}\)+\(\dfrac{2}{6.8}\)+...+\(\dfrac{2}{2008.2010}\))
F=2.(\(\dfrac{1}{2}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{8}\)+...+\(\dfrac{1}{2008}\)-\(\dfrac{1}{2010}\))
F=2.(\(\dfrac{1}{2}\)-\(\dfrac{1}{2010}\))
F=\(\dfrac{1004}{1005}\)
Ta có \(63,1.2-21,3.6=0,9.7.10.1,2-21.3,6\)
\(=6,3.1,2-21.3,6\)
\(=0,9.7.4.3-7.3.0,9.4\)
\(=6,3.1,2-6,3.1,2\)
\(=0\)
\(\Rightarrow\dfrac{\left(1+2+......+100\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+.....+99-100}=\dfrac{\left(1+2+.....+100\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{9}\right)0}{1-2+3-4+......+99-100}=0\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(B=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{107.111}\)
\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{107}-\frac{1}{111}\)
\(B=\frac{1}{3}-\frac{1}{111}\)
\(B=\frac{12}{37}\)
\(C=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(C=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(C=7\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(C=7.\frac{3}{35}\)
\(C=\frac{3}{5}\)
Ta có:
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{1}-\frac{1}{100}=\frac{99}{100}\)
\(B=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{107.111}\)
\(B=4.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}\right)\)
\(B=4.\left(\frac{1}{3}-\frac{1}{111}\right)=4.\frac{12}{37}=\frac{48}{37}\)
\(C=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(C=7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)
\(C=7.\left(\frac{1}{10}-\frac{1}{70}\right)=7.\frac{3}{35}=\frac{3}{5}\)
c: \(C=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{100}{99}=\dfrac{100}{2}=50\)
d: \(D=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot..\cdot\dfrac{98}{99}=\dfrac{1}{99}\)