Tính
\(1-\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56}-\frac{17}{72}+\frac{19}{90}\)
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ta có: \(1-\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56}-\frac{17}{72}+\frac{19}{90}\)
\(=1-\frac{5}{2.3}+\frac{7}{3.4}-\frac{9}{4.5}+\frac{11}{5.6}-\frac{13}{6.7}+\frac{15}{7.8}-\frac{17}{8.9}+\frac{19}{9.10}\)
\(=1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+\frac{1}{5}-\frac{1}{5}+\frac{1}{6}-\frac{1}{6}+\frac{1}{7}-\frac{1}{7}+\frac{1}{8}-\frac{1}{8}+\frac{1}{9}-\frac{1}{9}+\frac{1}{10}\)
\(=1-\frac{1}{2}+\frac{1}{10}=\frac{3}{5}\)
Đè thừa một số \(\frac{25}{156}\),mk ko lại đề bài nhé
\(A=1-\frac{2+3}{2\cdot3}+.....+\frac{11+12}{11\cdot12}-\frac{12+13}{12\cdot13}\)
\(=1-\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+\frac{1}{4}-...+\frac{1}{11}+\frac{1}{12}-\frac{1}{12}-\frac{1}{13}\)
\(=\frac{1}{2}-\frac{1}{13}=\frac{11}{26}\)
\(=\left(1+\frac{1}{2}\right)-1+\frac{1}{6}+\left(\frac{1}{2}+\frac{1}{12}\right)-\frac{1}{2}+\frac{1}{20}+\left(\frac{1}{3}+\frac{1}{30}\right)-\frac{1}{3}+\frac{1}{42}+\left(\frac{1}{4}+\frac{1}{56}\right)-\frac{1}{4}+\frac{1}{72}\)
=\(=\left(1-1+\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}\right)+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{72}\right)\)
\(=0+\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{8\cdot9}\right)=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}=\left(1-\frac{1}{9}\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+...+\left(\frac{1}{8}-\frac{1}{8}\right)\)\(=\left(\frac{9}{9}-\frac{1}{9}\right)+0+...+0=\frac{8}{9}\)
\(M=\left(\dfrac{3}{2}-\dfrac{5}{6}+\dfrac{7}{12}-\dfrac{17}{72}\right)+\left(-\dfrac{9}{20}+\dfrac{11}{30}\right)+\left(\dfrac{-13}{42}+\dfrac{15}{56}\right)\)
\(=\dfrac{108}{72}-\dfrac{60}{72}+\dfrac{42}{72}-\dfrac{17}{72}+\dfrac{-27}{60}+\dfrac{22}{60}+\dfrac{-52}{168}+\dfrac{45}{168}\)
\(=\dfrac{73}{72}-\dfrac{1}{12}-\dfrac{1}{24}=\dfrac{73}{72}-\dfrac{6}{72}-\dfrac{3}{72}=\dfrac{64}{72}=\dfrac{8}{9}\)