\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}=\dfrac{1}{2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\\ =\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\\ =\left(-\dfrac{1}{3}+\dfrac{1}{3}\right)+\left(-\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(-\dfrac{1}{5}+\dfrac{1}{5}\right)+\left(-\dfrac{1}{6}+\dfrac{1}{6}\right)+\left(-\dfrac{1}{7}+\dfrac{1}{7}\right)+\left(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{8}\right)\\ =\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{8}=1-\dfrac{1}{8}=\dfrac{8-1}{8}=\dfrac{7}{8}\)

`@` `\text {Ans}`

`\downarrow`

\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)

`=`\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+\dfrac{1}{7\times8}\)

`=`\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\)

`=`\(1-\dfrac{1}{8}\)

`=`\(\dfrac{7}{8}\)

sos 

a 25 phần 41

b 9

c 1 phần 12

d 12

e 14 phần 15

f 24 phần 7

a 25/41 

b 9

c 1/12

d 12

e 14/15

f24/7

A = \(\dfrac{1}{2}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\)+...+ \(\dfrac{1}{812}\) + \(\dfrac{1}{870}\)

A = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\)+...+ \(\dfrac{1}{28\times29}\)+ \(\dfrac{1}{29\times30}\)

A = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\)  - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) +...+\(\dfrac{1}{28}\)-\(\dfrac{1}{29}\)+ \(\dfrac{1}{29}\) - \(\dfrac{1}{30}\)

A = 1 - \(\dfrac{1}{30}\)

A = \(\dfrac{29}{30}\)

29/30 nha

`S=1/2 +1/6 +1/12 +1/20 +...+1/380`

`=1/(1.2)+1/(2.3) +1/(3.4)+1/(4.5)+...+1/(19.20)`

`=1-1/2 +1/2 -1/3 +1/3-1/4 +1/4 -1/5+....+1/19-1/20`

`=1-1/20=20/20 -1/20 =19/20`

a) /x-21/=5                                   C) chiu thoi

x-21=5 & x-21=-5

x=5+21    x=-5+21

x=26         x=16

b) x = -1 & x =2                

đây là toán lớp 1 hả????
 

=1/1x2+1/2x3+1/3x4+1/4x5+1/5x6+1/6x7

=1-1/7

=6/7

\(\frac{113}{140}\frac{ }{ }\)

=\(\frac{6}{7}\)

ta có :

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

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

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

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

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

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

a/  Tinh giá trị:

\(D=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{10}\right)\) \(\Leftrightarrow D=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{7}{8}.\frac{8}{9}.\frac{9}{10}=\frac{1}{10}\) 

b/  Chứng minh:

\(E=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2}\) 

-  Với mọi số tự nhiên n khác không thì luôn có:   \(\frac{1}{n^2}< \frac{1}{\left(n-1\right)\left(n+1\right)}=\frac{1}{2}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)\) Do đó:

 \(E=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}=\) 

   \(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...+\frac{1}{99}-\frac{1}{101}\right)\)\(=\frac{1}{2}\left(1-\frac{1}{101}\right)< \frac{1}{2}\) Vậy \(E< \frac{1}{2}\) 

c/  Chứng minh : \(F=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{199}+\frac{1}{200}>\frac{7}{12}\) 

    \(F=\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{150}\right)+\left(\frac{1}{151}+\frac{1}{152}+...+\frac{1}{200}\right)>\frac{50}{150}+\frac{50}{200}=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)

   Vậy:            \(F>\frac{7}{12}\) .