Sửa đề : \(A=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{120}\)

\(A=\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{240}\)

\(A=2.\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+...+\dfrac{1}{240}\right)\)

\(A=2.\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)\)

\(A=2.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)

\(A=2.\left(\dfrac{1}{4}-\dfrac{1}{16}\right)=2.\left(\dfrac{4}{16}-\dfrac{1}{16}\right)=2.\dfrac{3}{16}=\dfrac{3}{8}\)

7/48 - (1/2 x 2 + 1/6 x 4 + 1/8 x 5 + 1/12 x 7 + 1/14 x 8) : x = 0

7/48 - (1 + 2/3 + 5/8 + 7/12 + 4/7) : x = 0 (đã rút gọn)

7/48 - (336/336 + 224/336 + 210/336 + 196/336 + 192/336) : x = 0 (quy đồng)

7/48 - 193/56 : x  = 0

193/56 : x = 0 + 7/48

193/56 : x = 7/48

              x = 193/56 : 7/48

              x = 1158/49

Ta có:

\(\frac{1}{2^2}< \frac{1}{1\cdot2}\)

\(\frac{1}{3^2}< \frac{1}{2\cdot3}\)

.............

\(\frac{1}{10^2}< \frac{1}{9\cdot10}\)

Suy ra:

\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{10^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{9\cdot10}\)

Suy ra: \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{10^2}< \frac{9}{10}< 1\)

Vậy ...............

Giúp mình nhanh nha. Thanks các bạn 

Ta có : \(\frac{1}{2^2}+\frac{1}{3^2}+..+\frac{1}{10^2}=\left(\frac{1}{2^2}+\frac{1}{3^2}\right)+..+\left(\frac{1}{8^2}+\frac{1}{9^2}\right)+\frac{1}{10^2}< \frac{1}{2^2}.2+\frac{1}{4^2}.2+\frac{1}{6^2}.2+\frac{1}{8^2}.2+\frac{1}{10^2}\)

\(=\frac{1}{2}+\frac{1}{8}+\frac{1}{18}+\frac{1}{32}+\frac{1}{100}=\frac{5197}{7200}< 1\)

=>ĐPCM

Bài làm:

Ta có: \(\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{98.100}\)

\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\right)+\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{99}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(=\frac{1}{2}.\frac{98}{99}+\frac{1}{2}.\frac{49}{100}\)

\(=\frac{49}{99}+\frac{49}{200}\)

\(=\frac{14651}{19800}\)

giúp vs

= 1- 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +... + 1/99 - 1/100

= 1 - 99/100

= 1/100.

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

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

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

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

ta có :

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}=\frac{30+10+5+3+2}{60}=\frac{50}{60}=\frac{5}{6}\)

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)

\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+\frac{6-5}{5.6}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

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