\(\dfrac{1}{100.99}-\dfrac{1}{99.98}-...-\dfrac{1}{2.1}\)

\(=\dfrac{1}{99}-\dfrac{1}{100}-\dfrac{1}{98}+\dfrac{1}{99}-\dfrac{1}{97}+\dfrac{1}{98}-...-\dfrac{1}{2}+\dfrac{1}{3}-1+\dfrac{1}{2}\)

\(=\dfrac{2}{99}-\dfrac{1}{100}-1=-\dfrac{9799}{9900}\)

mãi mới thấy

cậu tách đi nhé

Ta có: \(A=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{1}{1\cdot99}+\dfrac{1}{3\cdot97}+\dfrac{1}{5\cdot95}+...+\dfrac{1}{97\cdot3}+\dfrac{1}{99\cdot1}}\)
\(\Leftrightarrow\dfrac{A}{100}=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{100}{1\cdot99}+\dfrac{100}{3\cdot97}+\dfrac{100}{5\cdot95}+...+\dfrac{100}{97\cdot3}+\dfrac{100}{99\cdot1}}\)

\(\Leftrightarrow\dfrac{A}{100}=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{97}+\dfrac{1}{99}}{1+\dfrac{1}{99}+\dfrac{1}{3}+\dfrac{1}{97}+\dfrac{1}{5}+\dfrac{1}{95}+...+\dfrac{1}{97}+\dfrac{1}{3}+\dfrac{1}{99}+1}\)

\(\Leftrightarrow\dfrac{A}{100}=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{97}+\dfrac{1}{99}}{2\left(1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{97}+\dfrac{1}{99}\right)}\)

\(\Leftrightarrow\dfrac{A}{100}=\dfrac{1}{2}\)

hay A=50

`4 1/5 xx 2 1/4`

`= 21/5 xx 9/4`

`= 189/20`

__

`4 1/5 : 2 1/4`

`= 21/5 : 9/4`

`= 21/5 xx 4/9`

`=84/45`

`=28/15`

__

`3 3/5 xx 1 2/3`

`= 18/5 xx 5/3`

`= 90/15`

`=6`

__

`3 3/5 : 1 2/3`

`= 18/5 : 5/3`

`= 18/5 xx 3/5`

`=54/25`

\(4\dfrac{1}{5}\times2\dfrac{1}{4}\\ =\dfrac{21}{5}\times\dfrac{9}{4}\\ =\dfrac{21\times9}{5\times4}\\ =\dfrac{189}{20}\)

\(3\dfrac{3}{5}\times1\dfrac{2}{3}\\ =\dfrac{18}{5}\times\dfrac{5}{3}\\ =\dfrac{18\times5}{5\times3}\\ =\dfrac{90}{15}\\ =6\)

\(4\dfrac{1}{5}:2\dfrac{1}{4}\\ =\dfrac{21}{5}:\dfrac{9}{4}\\ =\dfrac{21}{5}\times\dfrac{4}{9}\\ =\dfrac{21\times4}{5\times9}\\ =\dfrac{84}{45}\\ =\dfrac{28}{15}\)

\(3\dfrac{3}{5}:1\dfrac{2}{3}\\ =\dfrac{18}{5}:\dfrac{5}{3}\\ =\dfrac{18}{5}\times\dfrac{3}{5}\\ =\dfrac{18\times3}{5\times5}\\ =\dfrac{54}{25}\)

Đặt A = \(\dfrac{5}{2.1}+\dfrac{4}{1.11}+\dfrac{3}{11.2}+\dfrac{1}{2.15}+\dfrac{13}{15.4}\)

\(\dfrac{1}{7}A=\dfrac{1}{7}\left(\dfrac{5}{2.1}+\dfrac{4}{1.11}+\dfrac{3}{11.2}+\dfrac{1}{2.15}+\dfrac{13}{15.4}\right)\)

\(=\dfrac{5}{2.7}+\dfrac{4}{7.11}+\dfrac{3}{11.14}+\dfrac{1}{14.15}+\dfrac{13}{15.28}\)

\(=\dfrac{7-2}{2.7}+\dfrac{11-7}{7.11}+\dfrac{14-11}{11.14}+\dfrac{15-14}{14.15}+\dfrac{28-15}{15.28}\)

\(=\dfrac{7}{2.7}-\dfrac{2}{2.7}+\dfrac{11}{7.11}-\dfrac{7}{7.11}+\dfrac{14}{11.14}-\dfrac{11}{11.14}+\dfrac{15}{14.15}-\dfrac{14}{14.15}+\dfrac{28}{15.28}-\dfrac{15}{15.28}\)

\(=\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{28}\)

\(=\dfrac{1}{2}-\dfrac{1}{28}=\dfrac{14}{28}-\dfrac{1}{28}=\dfrac{13}{28}\)

\(A=\dfrac{13}{28}\div\dfrac{1}{7}=\dfrac{13}{4}\)

Đặt A = \(\dfrac{5}{2.1}+\dfrac{4}{1.11}+\dfrac{3}{11.2}+\dfrac{1}{2.15}+\dfrac{13}{15.4}\)

\(\Rightarrow\dfrac{1}{7}.A=\dfrac{5}{2.7}+\dfrac{4}{7.11}+\dfrac{3}{11.14}+\dfrac{1}{14.15}+\dfrac{13}{15.28}\)

\(\Rightarrow\dfrac{1}{7}.A=\left(\dfrac{1}{2}-\dfrac{1}{7}\right)+\left(\dfrac{1}{7}-\dfrac{1}{11}\right)+\left(\dfrac{1}{11}-\dfrac{1}{14}\right)+\left(\dfrac{1}{14}-\dfrac{1}{15}\right)+\left(\dfrac{1}{15}-\dfrac{1}{28}\right)\)

\(\Rightarrow\dfrac{1}{7}.A=\dfrac{1}{2}-\dfrac{1}{28}=\dfrac{13}{28}\)

\(\Leftrightarrow A=\dfrac{13}{4}\)

Vậy...................

a) \(\dfrac{9}{11}\times8=\dfrac{9\times8}{11}=\dfrac{72}{11}\)

b) \(\dfrac{4}{5}\times1=\dfrac{4\times1}{5}=\dfrac{4}{5}\)

c) \(\dfrac{15}{8}\times0=\dfrac{15\times0}{8}=\dfrac{0}{8}=0\)

a: 9/11*8=(9*8)/11=72/11

b: 4/5*1=(4*1)/5=4/5

c: 15/8*0=(15*0)/8=0/8=0

C= 1/100-(1/1.2+1/2.3+...+1/97.98+1/98.99+1/99.100)

C=1/100-(1-1/2+1/2-1/3+...+1/97-1/98+1/98-1/99+1/99-1/100)

C=1/100-(1-1/100)

C=1/100-99/100

C=-98/100=-49/50

\(C=\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-...\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

\(=-\left(\dfrac{1}{100.99}+\dfrac{1}{99.98}+\dfrac{1}{98.97}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)+\dfrac{1}{100}\)

\(=-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{97.98}+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)+\dfrac{1}{100}\)

\(=-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)+\dfrac{1}{100}\)

\(=-\left(1-\dfrac{1}{100}\right)+\dfrac{1}{100}\)

\(=\left(-1\right)+\dfrac{1}{50}=-\dfrac{49}{50}\)

theo bài ra ta có:

\(C=\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\\ \Rightarrow C=\dfrac{1}{100}-\left(\dfrac{1}{100.99}+\dfrac{1}{99.98}+\dfrac{1}{98.97}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)\\ \Rightarrow C=\dfrac{1}{100}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{97.98}+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\\ \Rightarrow C=\dfrac{1}{100}-\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\) \(\Rightarrow C=\dfrac{1}{100}-\left(\dfrac{1}{1}-\dfrac{1}{100}\right)\\ \Rightarrow C=\dfrac{1}{100}-\dfrac{99}{100}\\ \Rightarrow C=\dfrac{-98}{100}=\dfrac{-49}{50}\)