a)   \(D=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+...+\frac{1}{512}+\frac{1}{1024}\)

=>  \(2D=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...++\frac{1}{256}+\frac{1}{512}\)

=>  \(2D-D=\left(1+\frac{1}{2}+...+\frac{1}{512}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\right)\)

=>  \(D=1-\frac{1}{1024}\)

b)  \(Đ=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}+\frac{1}{19.20}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\)

\(=1-\frac{1}{20}=\frac{19}{20}\)

a) D=\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\dots+\frac{1}{512}+\frac{1}{1024}.\) 

\(D=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\dots+\frac{1}{512}-\frac{1}{1024}\)

\(D=1-\frac{1}{1024}\)

\(D=\frac{1023}{1024}\)

\(Đ=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\dots+\frac{1}{18\cdot19}+\frac{1}{19\cdot20}\)

\(Đ=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\dots+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\)

\(Đ=1-\frac{1}{20}\) 

\(Đ=\frac{19}{20}\)

Phần c như kiểu sai đề chỗ cuối hay sao ấy.

a)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{98.99}+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

b)\(\frac{5}{11.16}+\frac{5}{16.21}+...+\frac{5}{61.66}\)

\(=\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+....+\frac{1}{61}-\frac{1}{66}\)

\(=\frac{1}{11}-\frac{1}{66}\)

\(=\frac{5}{66}\)

a,\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)

ta có:

\(\frac{1}{1.2}=\frac{2-1}{1.2}=\frac{2}{1.2}-\frac{1}{1.2}=1-\frac{1}{2}\)

\(\frac{1}{2.3}=\frac{3-2}{2.3}=\frac{3}{2.3}-\frac{2}{2.3}=\frac{1}{2}-\frac{1}{3}\)

...

\(\frac{1}{99.100}=\frac{1}{99}-\frac{1}{100}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

=\(1-\frac{1}{100}=\frac{99}{100}\)

b,

\(\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.16}+...+\frac{5}{61.66}\)

ta có:

\(\frac{5}{11.16}=\frac{16-11}{11.16}=\frac{16}{11.16}-\frac{11}{11.16}=\frac{1}{11}-\frac{1}{16}\)

\(\frac{5}{16.21}=\frac{21-16}{16.21}=\frac{21}{16.21}-\frac{16}{16.21}=\frac{1}{16}-\frac{1}{21}\)

...

\(\frac{5}{61.66}=\frac{66-61}{61.66}=\frac{66}{61.66}-\frac{61}{61.66}=\frac{1}{61}-\frac{1}{66}\)

= \(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+...+\frac{1}{61}-\frac{1}{66}\)

=\(\frac{1}{11}-\frac{1}{66}\)=\(\frac{5}{66}\)

mik cũng đang ko biết bài này nè

mình cũng thế

1/6+1/12+1/20+...+1/110 = 

5B = 1/11 - 1/16 + 1/16 - 1/21 + 1/21 - 1/26 +.........+ 1/56 - 1/61 +1/61 - 1/66 
5B = 1/11 - 1/66 
5B = 5/66 
B = 1/66

1/66

nha bn

\(B=\frac{1}{11.16}+\frac{1}{16.21}+...+\frac{1}{61.66}\)

\(\Rightarrow5B=\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+...+\frac{1}{61}-\frac{1}{66}\)

\(\Rightarrow5B=\frac{6}{66}-\frac{1}{66}=\frac{5}{66}\)

\(\Rightarrow B=\frac{1}{66}\)

a, \(\frac{5}{11.16}+\frac{5}{11.21}+\frac{5}{21.26}+...+\frac{5}{61.66}\)

=\(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}+...+\frac{1}{61}-\frac{1}{66}\)

=\(\frac{1}{11}-\frac{1}{66}=\frac{5}{66}\)

a, A=1/11-1/16+1/16-1/21+1/21-1/26+...+1/61-1/66

      = 1/11-1/66=5/66 ( A chính là biểu thức ở phần a)

b, 1/12+1/20+1/30+...+1/110

=1/3.4+1/4.5+1/5.6+..+1/10.11

=1/3-1/4+1/4-1/5+1/5-1/6+...+1/10-1/11

=1/3-1/11=8/33

b,\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{10.11}\) 

=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)

=\(\frac{1}{2}-\frac{1}{11}=\frac{9}{22}\)

Ghi đầy đủ nha

bn có thể ghi rõ ràng đc ko?

a)\(-\dfrac{3}{7}+\dfrac{5}{13}+\dfrac{3}{7}\) 

=\(\left(-\dfrac{3}{7}+\dfrac{3}{7}\right)+\dfrac{5}{13}\) 

=\(0+\dfrac{5}{13}\) 

=\(\dfrac{5}{13}\)

a.\(\dfrac{27}{8}\)

b.\(\dfrac{37}{40}\)
c.\(\dfrac{5}{2}\)

d.\(\dfrac{7}{3}\)

e.5

g.\(\dfrac{53}{16}\)

Bài 1 :

a) \(\dfrac{3}{2}+\dfrac{5}{4}+\dfrac{5}{8}=\dfrac{12}{8}+\dfrac{10}{8}+\dfrac{5}{8}=\dfrac{12+10+5}{8}=\dfrac{27}{8}\)

b) \(\dfrac{4}{5}-\dfrac{3}{8}+\dfrac{2}{4}=\dfrac{32}{40}-\dfrac{15}{40}+\dfrac{20}{40}=\dfrac{32-15+20}{40}=\dfrac{37}{40}\)

c) \(3+\dfrac{6}{8}-\dfrac{5}{4}=\dfrac{3}{1}+\dfrac{6}{8}-\dfrac{5}{4}=\dfrac{24}{8}+\dfrac{6}{8}-\dfrac{10}{8}=\dfrac{20}{8}=\dfrac{5}{2}\)

d) \(\dfrac{5}{6}-\dfrac{1}{2}+2=\dfrac{5}{6}-\dfrac{1}{2}+\dfrac{2}{1}=\dfrac{5}{6}-\dfrac{3}{6}+\dfrac{12}{6}=\dfrac{14}{6}=\dfrac{7}{3}\)

e) \(\dfrac{3}{5}+\dfrac{6}{11}+\dfrac{7}{13}+\dfrac{2}{5}+\dfrac{16}{11}+\dfrac{19}{13}=\left(\dfrac{3}{5}+\dfrac{2}{5}\right)+\left(\dfrac{6}{11}+\dfrac{16}{11}\right)+\left(\dfrac{7}{13}+\dfrac{19}{13}\right)=1+2+2=5\)

g) \(\dfrac{75}{100}+\dfrac{18}{21}+\dfrac{29}{32}+\dfrac{1}{4}+\dfrac{3}{21}+\dfrac{13}{32}=\dfrac{3}{4}+\dfrac{6}{7}+\dfrac{29}{32}+\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{13}{32}=\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+\left(\dfrac{6}{7}+\dfrac{1}{7}\right)+\left(\dfrac{29}{32}+\dfrac{13}{32}\right)=1+1+\dfrac{21}{16}=2+\dfrac{21}{16}=\dfrac{53}{16}\)

a,3/2+5/4+5/8 = 12/8 + 10/8 + 5/8 = 27/8

b,4/5-3/8+2/4 = 32/40 - 15/40 + 20/40 = 37/40

c,3+6/8-5/4= 24/8 + 6/8 - 10/8 = 20/8 = 5/2

d,5/6-1/2+2 = 5/6 - 3/6 + 12/6 = 14/6 = 7/3

e,3/5+6/11+7/13+2/5+16/11+19/13 = ( 3/5+ 2/5) + ( 6/11 + 16/11) + ( 7/13 + 19/13) = 1 + 2 + 2 = 5

g,75/100+18/21+29/32+1/4+3/21+13/32 = 3/4 + 6/7 + 29/32 + 1/4 + 1/7 + 13/32=  ( 3/4 + 1/4) + ( 29/32 + 13/32) + ( 6/7 + 1/7)= 1 + 21/16 + 1 = 53/16

a,Ta có :

1/3 + 1/16 + 1/19 + 1/21 + 1/61 + 1/72 + 1/83 + 1/94 = 0,54

3/4 = 0,6

=>b < 3/4