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\(A=-\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}\)
\(=\frac{-6.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}{9.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}\)
\(=-\frac{6}{9}=-\frac{2}{3}\)
`Answer:`
\(C=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)...\left(1-\frac{1}{210}\right)\)
\(=\left(\frac{3}{3}-\frac{1}{3}\right)\left(\frac{6}{6}-\frac{1}{6}\right)\left(\frac{10}{10}-\frac{1}{10}\right)\left(\frac{15}{15}-\frac{1}{15}\right)...\left(\frac{210}{210}-\frac{1}{210}\right)\)
\(=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}...\frac{209}{210}\)
\(=\frac{2.2}{3.2}.\frac{5.2}{6.2}.\frac{9.2}{10.2}...\frac{209.2}{210.2}\)
\(=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}...\frac{418}{420}\)
\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{19.22}{20.21}\)
\(=\frac{1.4.2.5.3.6...19.22}{2.3.3.4.4.5...20.21}\)
\(=\frac{\left(1.2.3...19\right)\left(4.5.6...22\right)}{\left(2.3.4...20\right)\left(3.4.5...21\right)}\)
\(=\frac{11}{30}\)
\(P=\frac{1}{2000.1999}+\frac{1}{1999.1998}+...+\frac{1}{3.2}+\frac{1}{2.1}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1998.1999}+\frac{1}{1999.2000}\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1998}-\frac{1}{1999}+\frac{1}{1999}-\frac{1}{2000}\)
\(=\frac{1}{2}-\frac{1}{2000}=\frac{999}{2000}\)
\(P=\frac{1}{2000.1999}+\frac{1}{1999.1998}+..+\frac{1}{3.2}+\frac{1}{2.1}\)
=\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1998.1999}+\frac{1}{1999.2000}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..+\frac{1}{1999}-\frac{1}{2000}\)
=\(1-\frac{1}{2000}\)
=\(\frac{1999}{2000}\)
\(A=\left(\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\right)-\left(\frac{1}{3}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)
\(3^2A=3^2\left(\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\right)-3^2\left(\frac{1}{3}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)
\(9A=\left(1+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(3+\frac{1}{3}+...+\frac{1}{3^{97}}\right)\)
\(9A-A=\left(1-\frac{1}{3^{100}}\right)-\left(3-\frac{1}{3^{99}}\right)\)
\(8A=1-3=-2\)
A=\(\frac{-2}{8}=\frac{-1}{4}\)
\(B=4\left|\frac{-1}{4}\right|+\frac{1}{3^{100}}=1+\frac{1}{3^{100}}=1\)
Vậy B=1
2Q = 1-1/3-1/2+1/4+1/3-1/5-1/4+1/6-........+1/97-1/99-1/98+1/100 = 1-1/2-1/99+1/100 = 4949/9900 >> Q = 49499/19800
\(Q=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+...+\frac{1}{97.99}-\frac{1}{98.100}\)
\(=\frac{1}{2}\left(1-\frac{1}{3}-\frac{1}{2}+\frac{1}{4}+\frac{1}{3}+\frac{1}{5}-\frac{1}{4}+\frac{1}{6}+...+\frac{1}{97}-\frac{1}{99}-\frac{1}{98}+\frac{1}{100}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{100}\right)=\frac{1}{2}.\frac{99}{100}=\frac{99}{200}\) (không chắc cho lắm :v)
=1-(1/5+1/5-1/4+1/4+....+1/101-1/52)
=1-1/52
=51/52