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BT
22 tháng 3 2018
Ta có: \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{19}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{20}\right)\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{19}\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{20}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{20}\right)=\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{19}+\frac{1}{20}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\right)=\)
= 1/11 + 1/12 +1/13+...+1/20 (đpcm)
AC
1
MN
15 tháng 3 2016
\(\left(-1\frac{1}{6}\right)\left(\frac{1-\frac{3}{5}+\frac{3}{11}-\frac{3}{13}}{\frac{1}{3}-\frac{1}{5}+\frac{1}{11}-\frac{1}{13}}\right)\left(\frac{4-\frac{4}{17}+\frac{4}{19}-\frac{4}{2013}}{5-\frac{5}{7}+\frac{5}{19}-\frac{5}{2013}}\right)\)
\(=-\frac{7}{6}.\left(\frac{3\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{11}-\frac{1}{13}\right)}{\frac{1}{3}-\frac{1}{5}+\frac{1}{11}-\frac{1}{13}}\right):\left(\frac{4.\left(1-\frac{1}{7}+\frac{1}{19}-\frac{1}{2013}\right)}{5.\left(1-\frac{1}{7}+\frac{1}{19}-\frac{1}{2013}\right)}\right)\)
\(=-\frac{7}{6}.3:\frac{4}{5}=-\frac{7}{2}.\frac{5}{4}=-\frac{35}{8}\)
T
30 tháng 4 2019
1/ Tính:
\(\frac{3}{2}-\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56}-\frac{17}{72}+\frac{19}{90}\)
\(=\frac{3}{1.2}-\frac{5}{2.3}+\frac{7}{3.4}-\frac{9}{4.5}+\frac{11}{5.6}-\frac{13}{6.7}+\frac{15}{7.8}-\frac{17}{8.9}+\frac{19}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
KC
23 tháng 4 2018
\(\frac{\frac{4}{17}+\frac{4}{19}-\frac{4}{2111}}{\frac{5}{17}+\frac{5}{19}-\frac{5}{2111}}-\frac{\frac{1}{123}-\frac{1}{19}+\frac{1}{371}-\frac{1}{5}}{-\frac{5}{123}+\frac{5}{19}-\frac{5}{371}+1}\)
\(=\frac{4.\left(\frac{1}{17}+\frac{1}{19}-\frac{1}{2111}\right)}{5.\left(\frac{1}{17}+\frac{1}{19}-\frac{1}{2111}\right)}+\frac{\frac{1}{123}-\frac{1}{19}+\frac{1}{371}-\frac{1}{5}}{5.\left(\frac{1}{123}-\frac{1}{19}+\frac{1}{371}-\frac{1}{5}\right)}=\frac{4}{5}+\frac{1}{5}=1\)
29 tháng 6 2018
Cho tam giác ABC có đường cao AD .Gọi E là trung điểm của AB .F đối xứng vs D qua E c/m AB = DF
18 tháng 8 2020
[\(\frac{-75}{59}\).\(\frac{-107}{93}\)]\(\frac{31}{50}\)=\(\frac{2675}{1829}\).\(\frac{31}{50}\)=\(\frac{107}{118}\)
18 tháng 8 2020
\(\left[\frac{1\frac{11}{31}\cdot4\frac{3}{7}-\left(15-6\frac{1}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-5\frac{1}{3}\right)}\cdot\left(-1\frac{14}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\frac{\frac{42}{31}\cdot\frac{31}{7}-\left(15-\frac{19}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-\frac{16}{3}\right)}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\frac{6-\left(15-\frac{2}{3}\right)}{\frac{29}{6}+\frac{1}{6}\cdot\frac{20}{3}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\frac{6-15+\frac{2}{3}}{\frac{29}{6}+\frac{10}{9}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\frac{-\frac{25}{3}}{\frac{107}{18}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\left(-\frac{150}{107}\right)\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}=\frac{50}{31}\cdot\frac{31}{50}=1\)
VA
16 tháng 7 2016
\(A=\frac{636363.37-373737.63}{1+2+3+...+2006}\)
\(A=\frac{63.10101.37-37.10101.63}{1+2+3+...+2006}\)
\(A=0\)
\(B=1\frac{6}{41}.\left(\frac{12+\frac{12}{19}-\frac{12}{37}-\frac{12}{53}}{3+\frac{3}{19}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2006}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2006}}\right).\frac{124242423}{237373735}\)
\(B=\frac{47}{41}.\left[\frac{4.\left(3+\frac{3}{19}-\frac{3}{37}-\frac{3}{53}\right)}{1.\left(3+\frac{3}{19}-\frac{3}{37}-\frac{3}{53}\right)}:\frac{4.\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2006}\right)}{5.\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2006}\right)}\right].\frac{124242423}{237373735}\)
\(B=\frac{47}{41}.\left(4:\frac{4}{5}\right).\frac{124242423}{237373735}\)
\(B=\frac{47}{41}.5.\frac{124242423}{237373735}\)
\(B=\frac{47.5.124242423}{41.237373735}\)
\(B=\frac{29196969405}{9732323135}\)
Ủng hộ mk nha !!! ^_^
16 tháng 7 2016
a) \(A=\frac{636363.37-373737.63}{1+2+3+...+2006}\)
\(A=\frac{10101.63.37-10101.37.63}{1+2+3+...+2006}\)
\(A=\frac{0}{1+2+3+...+2006}\)
\(A=0\)
b) \(B=1\frac{6}{41}\left(\frac{12+\frac{12}{19}-\frac{12}{37}-\frac{12}{53}}{3+\frac{3}{19}-\frac{3}{37}-\frac{3}{53}}.\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2006}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2006}}\right).\frac{124242423}{237373735}\)
\(B=\frac{47}{41}.\frac{12}{3}.\left(\frac{1+\frac{1}{19}-\frac{1}{37}-\frac{1}{53}}{1+\frac{1}{19}-\frac{1}{37}-\frac{1}{53}}\right).\frac{4}{5}.\left(\frac{1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2006}}{1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2006}}\right).\frac{123}{235}\)
\(B=\frac{47.4.4.123}{41.5.235}\)
\(B=\frac{47.4.4.41.3}{41.5.47.5}\)
\(B=\frac{4.4.3}{5.5}\)
\(B=\frac{48}{25}\)
Vì \(\frac{1}{4}>\frac{1}{16};\frac{1}{5}>\frac{1}{16};...;\frac{1}{19}>\frac{1}{16}\)
\(\Rightarrow\)\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+.....+\frac{1}{19}>\frac{1}{16}+\frac{1}{16}+.....+\frac{1}{16}\) ( 16 số)
\(=\frac{1+1+1+.....+1}{16}\)
\(=\frac{16}{16}=1\)
Vậy: \(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+.....+\frac{1}{19}>1\)
\(\frac{1}{13}x14=\frac{14}{13}>1\)