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a: \(B=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2007}-\dfrac{1}{2008}=1-\dfrac{1}{2008}=\dfrac{2007}{2008}\)
b: \(Q=\dfrac{7}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2009\cdot2011}\right)\)
\(=\dfrac{7}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\right)\)
\(=\dfrac{7}{2}\cdot\dfrac{2010}{2011}\simeq3,50\)
A = \(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
A=\(\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{50}{100}-\dfrac{1}{100}=\dfrac{49}{100}\)
B = \(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{49.51}\)
B = \(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{49}-\dfrac{1}{51}\)
B = \(\dfrac{1}{2}-\dfrac{1}{51}=\dfrac{51}{102}-\dfrac{2}{102}=\dfrac{49}{102}\)
A = \(\dfrac{9}{1.2}\)+ \(\dfrac{9}{2.3}\)+\(\dfrac{9}{3.4}\)+......+\(\dfrac{99}{99.100}\)
A = 9( \(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+.......+\(\dfrac{1}{99.100}\))
A = 9( 1-\(\dfrac{1}{2}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+........+\(\dfrac{1}{99}\)-\(\dfrac{1}{100}\))
A = 9 ( 1 - \(\dfrac{1}{100}\))
A = 9 . \(\dfrac{99}{100}\)
A = \(\dfrac{891}{100}\)
\(A=\dfrac{9}{1\cdot2}+\dfrac{9}{2\cdot3}+\dfrac{9}{3\cdot4}+...+\dfrac{9}{98\cdot99}+\dfrac{9}{99\cdot100}\)
\(=9\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{98\cdot99}+\dfrac{1}{99\cdot100}\right)\)
\(=9\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=9\left(1-\dfrac{1}{100}\right)\)
\(=9\left(\dfrac{100}{100}-\dfrac{1}{100}\right)\)
\(=9\cdot\dfrac{99}{100}\)
\(=\dfrac{891}{100}\)
\(A=\dfrac{6}{7}+\dfrac{1}{7}.\dfrac{2}{7}+\dfrac{1}{7}.\dfrac{5}{7}.\)
\(A=\dfrac{6}{7}+\dfrac{1}{7}\left(\dfrac{2}{7}+\dfrac{5}{7}\right).\)
\(A=\dfrac{6}{7}+\dfrac{1}{7}.1.\)
\(A=\dfrac{6}{7}+\dfrac{1}{7}=1.\)
Vậy \(A=1.\)
\(B=\dfrac{40}{9}.\dfrac{13}{3}-\dfrac{4}{3}.\dfrac{40}{9}.\)
\(B=\dfrac{4}{9}.\dfrac{13}{3}-\dfrac{4}{9}.\dfrac{40}{3}.\)
\(B=\dfrac{4}{9}\left(\dfrac{13}{3}-\dfrac{40}{3}\right).\)
\(B=\dfrac{4}{9}.\left(-9\right).\)
\(B=-4.\)
Vậy \(B=-4.\)
F=(9.75.21\(\dfrac{3}{7}\)+\(\dfrac{39}{4}\).18\(\dfrac{4}{7}\)).\(\dfrac{15}{78}\)
=(\(\dfrac{39}{4}\).21\(\dfrac{3}{7}\)+\(\dfrac{39}{4}\).18\(\dfrac{4}{7}\)).\(\dfrac{15}{78}\)
=[\(\dfrac{39}{4}\).(21\(\dfrac{3}{7}\)+18\(\dfrac{4}{7}\))].\(\dfrac{15}{78}\)
=[\(\dfrac{39}{4}\).(21+18)+(\(\dfrac{3}{7}\)+\(\dfrac{4}{7}\))].\(\dfrac{15}{78}\)
=[\(\dfrac{39}{4}\).(39+1)].\(\dfrac{15}{78}\)
=(\(\dfrac{39}{4}\).40).\(\dfrac{15}{78}\)
=390.\(\dfrac{15}{78}\)=75
\(B=71\dfrac{38}{45}-\left(43\dfrac{8}{45}-1\dfrac{17}{57}\right)\)
\(B=71\dfrac{38}{45}-43\dfrac{8}{45}-1\dfrac{17}{57}\)
\(B=28\dfrac{2}{3}-1\dfrac{17}{57}=27\dfrac{11}{57}\)
\(D=\left(19\dfrac{5}{8}:\dfrac{7}{12}-13\dfrac{1}{4}:\dfrac{7}{12}\right).\dfrac{4}{5}\)
\(D=\dfrac{12}{7}.\left(19\dfrac{5}{8}-13\dfrac{1}{4}\right).\dfrac{4}{5}\)
\(D=\dfrac{12}{7}.\dfrac{51}{8}.\dfrac{4}{5}=\dfrac{306}{35}\)
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\(D=\left(\dfrac{157}{8}:\dfrac{7}{12}-\dfrac{53}{4}:\dfrac{7}{12}\right).\dfrac{4}{5}=\left(\dfrac{157.3.4}{2.4.7}-\dfrac{53.3.4}{4.7}\right).\dfrac{4}{5}=\left(\dfrac{417}{14}-\dfrac{159}{7}\right).\dfrac{4}{5}=\dfrac{153}{14}.\dfrac{4}{5}=\dfrac{306}{35}\)
a) \(\dfrac{3}{5}+0,145-\dfrac{1}{200}\)
\(=\dfrac{3}{5}+\dfrac{145}{1000}-\dfrac{1}{200}\)
\(=\dfrac{3}{5}+\dfrac{29}{200}-\dfrac{1}{200}\)
\(=\dfrac{120}{200}+\dfrac{29}{200}-\dfrac{1}{200}\)
\(=\dfrac{148}{200}\)
\(=\dfrac{37}{50}\)
b) \(\left(31\dfrac{6}{13}+5\dfrac{9}{41}\right)-36\dfrac{6}{13}\)
\(=31\dfrac{6}{13}+5\dfrac{9}{41}-36\dfrac{6}{13}\)
\(=\left(31\dfrac{6}{13}-36\dfrac{6}{13}\right)+5\dfrac{5}{41}\)
\(=\left(-5\right)+5\dfrac{5}{41}\)
\(=0\dfrac{5}{41}\)
\(=\dfrac{5}{41}\)
c) \(5.2\dfrac{1}{7}+5.7\dfrac{6}{7}\)
\(=5\left(2\dfrac{1}{7}+7\dfrac{6}{7}\right)\)
\(=5\left(9+\dfrac{1}{7}+\dfrac{6}{7}\right)\)
\(=5\left(9+1\right)\)
\(=5.10\)
\(=50\)
a) \(\dfrac{3}{5}+0,415-\dfrac{1}{200}\)
\(=\dfrac{3}{5}+\dfrac{83}{200}-\dfrac{1}{200}\\ =\dfrac{120}{200}+\dfrac{83}{200}-\dfrac{1}{200}\\ =\dfrac{120+83-1}{200}=\dfrac{202}{200}=\dfrac{101}{100}\)
b)\(\left(31\dfrac{6}{13}+5\dfrac{9}{41}\right)-36\dfrac{6}{13}\)
\(=\left(\dfrac{409}{13}+\dfrac{214}{41}\right)-\dfrac{474}{13}\)
\(=\dfrac{19551}{533}-\dfrac{474}{13}=\dfrac{9}{41}\)
c)\(5.2\dfrac{1}{7}+5.7\dfrac{6}{7}\)
\(=5.\dfrac{15}{7}+5.\dfrac{55}{7}\\ =5\left(\dfrac{15}{7}+\dfrac{55}{7}\right)\\ =5.10=50\)
1)Tính
a)\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+..........+\dfrac{1}{9.10}\)
=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}\)
\(=\dfrac{9}{10}\)
b)\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.........+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+..............+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
2) tìm x
\(a\)) \(\dfrac{2}{5}+\dfrac{4}{5}x-\dfrac{7}{5}\)\(=\dfrac{9}{5}\)
\(\dfrac{4}{5}x+\dfrac{7}{5}=\dfrac{9}{5}-\dfrac{2}{5}\)
\(\dfrac{4}{5}x+\dfrac{7}{5}=\dfrac{7}{5}\)
\(\dfrac{4}{5}x=\dfrac{7}{5}-\dfrac{7}{5}\)
\(\dfrac{4}{5}x=0\)
\(x=0:\dfrac{4}{5}\)
\(x=0\)
b)\(\dfrac{2}{5}x-\dfrac{6}{4}=\dfrac{8}{5}\)
\(\dfrac{2}{5}x=\dfrac{8}{5}+\dfrac{6}{4}\)
\(\dfrac{2}{5}x=\dfrac{31}{10}\)
\(x=\dfrac{31}{10}:\dfrac{2}{5}\)
\(x=\dfrac{31}{4}\)
1. Tính:
a. \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}\)
= \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
= \(\dfrac{1}{1}-\dfrac{1}{10}\)
= \(\dfrac{10}{10}-\dfrac{1}{10}\)
= \(\dfrac{9}{10}\)
b. \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
= \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
= \(\dfrac{1}{1}-\dfrac{1}{100}\)
= \(\dfrac{100}{100}-\dfrac{1}{100}\)
= \(\dfrac{99}{100}\)
2. Tìm x, biết:
a. \(\dfrac{2}{5}+\dfrac{4}{5}x-\dfrac{7}{5}=\dfrac{9}{5}\)
\(\dfrac{4}{5}x-\dfrac{7}{5}=\dfrac{9}{5}-\dfrac{2}{5}\)
\(\dfrac{4}{5}x-\dfrac{7}{5}=\dfrac{7}{5}\)
\(\dfrac{4}{5}x=\dfrac{7}{5}+\dfrac{7}{5}\)
\(\dfrac{4}{5}x=\dfrac{14}{5}\)
\(x=\dfrac{14}{5}:\dfrac{4}{5}\)
\(x=\dfrac{14}{5}.\dfrac{5}{4}\)
\(x=14.\dfrac{1}{4}\)
\(x=\dfrac{14}{4}\)
Vậy \(x=\dfrac{14}{4}\)
b. \(\dfrac{2}{5}x-\dfrac{6}{4}=\dfrac{8}{5}\)
\(\dfrac{2}{5}x=\dfrac{8}{5}+\dfrac{6}{4}\)
\(\dfrac{2}{5}x=\dfrac{32}{20}+\dfrac{30}{20}\)
\(\dfrac{2}{5}x=\dfrac{62}{20}\)
\(\dfrac{2}{5}x=\dfrac{31}{10}\)
\(x=\dfrac{31}{10}:\dfrac{2}{5}\)
\(x=\dfrac{31}{10}.\dfrac{5}{2}\)
\(x=\dfrac{31}{2}.\dfrac{2}{2}\)
\(x=\dfrac{31}{2}.1\)
\(x=\dfrac{31}{2}\)
Vậy \(x=\dfrac{31}{2}\)
bài này mk tự làm ko sao chép trên mạng
nếu thấy đúng thì tick đúng cho mk nha
\(A=\dfrac{7}{1.2}+\dfrac{7}{2.3}+\dfrac{7}{3.4}+...+\dfrac{7}{2011.2012}\)
\(A=7\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2011.2012}\right)\)
\(A=7\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2011}-\dfrac{1}{2012}\right)\)
\(A=7\left(1-\dfrac{1}{2012}\right)=7.\dfrac{2011}{2012}=\dfrac{14077}{2012}\)