K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

AH
Akai Haruma
Giáo viên
13 tháng 6 2022

Lời giải:

$\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}$

$=1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}$

$=(1+1+....+1)-(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90})$

$=8-(\frac{1}{2\times 3}+\frac{1}{3\times 4}+\frac{1}{4\times 5}+...+\frac{1}{9\times 10})$

$=8-(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10})$

$=8-(\frac{1}{2}-\frac{1}{10})=8-\frac{2}{5}=\frac{38}{5}$

21 tháng 7 2017

\(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+....+\frac{89}{90}\)

\(S=\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+\left(1-\frac{1}{20}\right)+....+\left(1-\frac{1}{90}\right)\)

\(S=\left(1-\frac{1}{2.3}\right)+\left(1-\frac{1}{3.4}\right)+\left(1-\frac{1}{4.5}\right)+....+\left(1-\frac{1}{9.10}\right)\)

\(S=\left(1+1+1+....+1\right)-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)

\(S=8-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(S=8-\left(\frac{1}{2}-\frac{1}{10}\right)\)

\(S=8-\frac{2}{5}=\frac{38}{5}\)

23 tháng 4 2016

1/2+5/6+11/12+19/20+...+89/90

                    Đ/s: 81/10.

23 tháng 4 2016

Đặt A = 1/2 + 5/6 + 11/12 + 19/20 + ... + 89/90

A = ( 1 - 1/2 ) + ( 1 - 1/12 ) + ( 1 - 1/20 ) + ... + ( 1 - 1/90 )

A = ( 1 + 1 + 1 + ... + 1 + 1 ) - ( 1/2 + 1/6 + 1/20 + ... + 1/90

A = 9 - ( 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/9.10 )

A = 9 - ( 1 - 1/10 )

A = 9 - 9/10

A = 81/10

bạn xem  lại đề bài vs ạ 91/2 hay là 1/2 ạ

24 tháng 5 2021

bạn ơi bn xem lại gíup mk 91/2 hay 1/2 tớ mới giải đc

DD
2 tháng 6 2021

a) \(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)

\(=1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+1-\frac{1}{30}+1-\frac{1}{42}+1-\frac{1}{56}+1-\frac{1}{72}+1-\frac{1}{90}\)

\(=8-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)

\(=8-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)

\(=8-\left(\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+\frac{6-5}{5.6}+\frac{7-6}{6.7}+\frac{8-7}{7.8}+\frac{9-8}{8.9}+\frac{10-9}{9.10}\right)\)

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

\(=8-\left(\frac{1}{2}-\frac{1}{10}\right)=7,6\)

b) Bạn làm tương tự. 

Ta có: \(\dfrac{1}{2}+\dfrac{5}{6}+\dfrac{11}{12}+\dfrac{19}{20}+\dfrac{41}{42}+\dfrac{55}{56}+\dfrac{71}{72}+\dfrac{89}{90}\)

\(=8-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\)

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

\(=8-1+\dfrac{1}{10}\)

\(=\dfrac{71}{10}\)

12 tháng 6 2016

Xem ở đây bạn nhé! http://olm.vn/hoi-dap/question/602318.html

S= 81/10.

1/2 + 5/6 + 11/12 + ... + 71/72 + 89/90 = 9 - ( 1/1x2 + 1/2x3 + ... + 1/9x10 ) = 9 - ( 1/1 - 1/2 + 1/2 - ... + 1/9 - 1/10 = 9 - ( 1 - 1/10 ) = 8 + 1/10 = 81/10

16 tháng 3 2017

\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+.....+\frac{71}{72}+\frac{89}{90}\)

\(=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+....+\left(1-\frac{1}{72}\right)+\left(1-\frac{1}{90}\right)\)

\(=\left(1+1+1+....+1\right)-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{90}\right)\)

\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{9.10}\right)\)

\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{9}-\frac{1}{10}\right)\)

\(=9-\left(1-\frac{1}{10}\right)=9-\frac{9}{10}=\frac{81}{10}\)