misa

Đặt tên biểu thức là A ta có :

\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+....+\frac{1}{90}+\frac{1}{110}\)

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

\(=\left(\frac{1}{2}-\frac{1}{11}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+.....+\left(\frac{1}{10}-\frac{1}{10}\right)\)

\(=\left(\frac{1}{2}-\frac{1}{11}\right)+0+......+0\)

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

\(\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{10.11}\)

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

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

=\(\frac{9}{22}\)

\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}+\frac{1}{110}\)

\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}+\frac{1}{10.11}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)

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

\(=\frac{10}{11}\)

   \(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{90}+\frac{1}{110}\)

\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{9.10}+\frac{1}{10.11}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.......+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)

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

\(=\frac{9}{22}\)

\(=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{10\cdot11}\)

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

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

\(=\frac{9}{22}\)

\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}+\frac{1}{110}\)

= \(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}+\frac{1}{10.11}\)

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

= \(\frac{1}{2}-\frac{1}{11}+0+...+0\)

= \(\frac{9}{22}\)

* Bạn tham khảo nhé *

1212 ++ 1616 ++ 112112 ++ 120120 ++ 130130 ++ 142142 ++ 156156

== 11×211×2 ++ 12×312×3 ++ 13×413×4 ++ 14×514×5 ++ 15×615×6 ++ 16×716×7 ++ 17×817×8

== 1111 −− 1212 ++ 1212 −− 1313 ++ 1313 −− 1414 ++ 1414 −− 1515 ++ 1515 −− 1616 ++ 1616 −− 1717 ++ 1717 −− 1818

== 1111 −− 1818

== 8888 −− 1818

== 78

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

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

\(=\dfrac{1}{2}-\dfrac{1}{6}\)

\(=\dfrac{1}{3}\)

\(\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}-\frac{1}{3\cdot4}-\frac{1}{4\cdot5}-\frac{1}{5\cdot6}\)

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

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

NHỚ K MK NHA. CHÚC BẠN HỌC TỐT

  1/2 - 1/6 - 1/12 - 1/20 - 1/30

=1/1x2 - 1/2x3- 1/3x4 - 1/4x5 - 1/5x6

=1-1/2 + 1/2-1/3 + 1/3-1/4 + 1/4-1/5 +1/5-1/6

=1-1/6

=5/6

Bài 1 : 

a) Hai phân số có chung tử số thì ta so sánh mẫu nếu mẫu lớn hơn thì phân số đó bé hơn 

Áp dụng vào đó ta có : 71 < 72 => 15/71 > 15/72

b) Ta có : 21/42 = 1/2 = 23/46 

Áp dụng câu a ta có : 46 > 45 => 21/42 < 23/45

c) Ta có : 47/45 = 1 + 2/45   ;  48/46 = 1 + 2/46

Vì 2/45 > 2/46 => 47/45 > 48/46 

d) Ta có : 1 - 13/25 = 12/25

1/3 = 12/36

Vì 12/25 > 12/36 => 13/25 > 3/7

Bài 2 :

D = 1/2 + 1/6 + 1/12 + 1/20 + ... + 1/110

D = 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + .... + 1/10.11

D = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + .... + 1/10 - 1/11

D = 1 - 1/11

D = 10/11

\(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{72}+\dfrac{1}{90}\)

\(=\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+...+\dfrac{1}{8\times9}+\dfrac{1}{9\times10}\)

\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=\dfrac{1}{3}-\dfrac{1}{10}\)

`=7/30`

help me

\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)

\(=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=1-\dfrac{1}{10}\)

\(=\dfrac{9}{10}\)

\((\frac{1}{2}+\frac{1}{4}+\frac{1}{6}) \)x \((\frac{36}{9}-\frac{12}{9}:\frac{1}{3}) \)

\(= (\frac{1}{2}+\frac{1}{4}+\frac{1}{6}) \) x \((\frac{36}{9}-\frac{36}{9})\)

\( = (\frac{1}{2}+\frac{1}{4}+\frac{1}{6}) \) x \(0\)

\(= 0\)