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

= 1 - 1/6 + 1 - 1/12 + 1 - 1/20 + ... + 1-1/90

= [1+1+1+1...+1] - [1/2*3 + 1/3*4 + 1/4*5 + ... +1/9*10]

= 8 - [1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/9 - 1/10]

= 8 - [1/2 - 1/10]

= 8 - 2/5

= 38/5

S=1-1/6+1-1/12+...+1-1/90

=8-(1/2.3+1/3.4+...+1/9.10)

=8-(1/2-1/3+1/3-1/4+...+1/9-1/10)

=8-(1/2-1/10)

=8-2/5

=38/5

S=5/6+11/12+19/20+...+89/90

=1-1/6+1-1/12+1-1/20+...+1-1/90

=(1+1+1+...+1)-(1/6+1/12+1/20+...+1/90)

=8-(1/2.3+1/3.4+1/4.5+...+1/9.10)

=8-[(3-2)/2.3+(4-3)/3.4+(5-4)/4.5+...+(10-9)/9.10]

=8-(3/2.3-2/2.3+4/3.4-3/3.4+5/4.5-4/4.5+...+10/9.10-10/9.10]

=8-(1/2-1/3+1/3-1/4+1/4-1/5+...+1/9-1/10)

=8-(1/2-1/10)

=8-2/5

=38/5

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

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

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

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

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

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

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

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

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

\(=\left(1+1+1+...+1\right)-\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\right)\)

Từ 2 đến 9 có : ( 9 - 2 ) / 1 + 1 = 8 ( số hạng ) => có 8 số 1

\(\Rightarrow8-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)

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

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

b) \(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+...+\frac{109}{110}\)

\(=1-\frac{1}{2}+1-\frac{1}{6}+...+1-\frac{1}{110}\)

\(=\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)\)

\(=\left(1+1+...+1\right)-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{10\cdot11}\right)\)

Từ 1 đến 10 có : ( 10 - 1 ) / 1 + 1 = 10 ( số hạng ) => có 10 số 1

\(\Rightarrow10-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)\)

\(=10-\left(1-\frac{1}{11}\right)\)

\(=10-\frac{10}{11}=\frac{100}{11}\)

100/11 nha

câu b bằng 891/110

câu a bằng 76/10

\(=\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{6}\right)+\left(1-\dfrac{1}{12}\right)+...+\left(1-\dfrac{1}{90}\right)\\ =\left(1+1+1+1+1+1+1+1+1\right)-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{90}\right)\\ =9-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{9\cdot10}\right)\\ =9-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ =9-\left(1-\dfrac{1}{10}\right)=9-\dfrac{9}{10}=\dfrac{81}{10}\)

81/10

8,1 nha

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

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

\(=8-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)

\(=8-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)

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

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

\(=\frac{38}{5}\)

100/11

hok tốt

`Answer:`

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

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

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

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

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

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

\(=\frac{38}{5}\)

b) \(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)

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

\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)

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

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

\(=\frac{81}{10}\)

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

                    Đ/s: 81/10.

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

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

nếu mà 1/2 thì tớ giải

9/10 nhe 

Bạn ra câu đố mik xin trả lời như sau :

Ta có : A = \(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)

A có số các phân số là :

          ( 89 - 5 ) : 1 + 1 = 85 ( p/số )

Ta có : A = \(\left(1-\frac{5}{6}\right)+\left(1-\frac{1}{12}\right)+\left(1-\frac{1}{20}\right)+...+\left(1-\frac{1}{90}\right)\)

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

A = \(\left(1.85\right)-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)( Có 85 số 1 vì mỗi số 1 đi kèm với 1 phân số mà có 85 phân số)

A = \(85-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)

A = \(85-\left(\frac{1}{2}-\frac{1}{10}\right)\)

A = \(85-\frac{2}{5}\)

A = 85 - 0.4

A = 84,6

Ủng hộ mik nhé ^_^"

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

= 1 - 1/6 + 1-1/12 + 1-1/20 + ... + 1-1/90

= 1 - 1/2*3 + 1 - 1/3*4 + 1 - 1/4*5 + ... + 1- 1/9*10

= [1 + 1 + 1 +... + 1] - [1/2*3 + 1/3*4 + 1/4*5 + ... + 1/9*10]

= 8 - [1/2 - 1/3 + 1/3 - 14 + 1/4 - 1/5 + ... + 1/9 - 1/10]

= 8 - [1/2 - 1/10]

= 8 - 2/5

= 38/5

A = (1 -1/2) + (1 - 1/6) + (1 - 1/12)  + (1 - 1/20 ) + ...+ (1 - 1/ 90)

= (1+1+1+1+1+1+1+1+1) - ( 1/2 - 1/6 - 1/12 - 1/ 20 - ...- 1/90)\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)\(=9-\left(1-\frac{1}{10}\right)=\frac{81}{10}\)

dap an dung la 81/10

\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)

\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)

\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)

\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)

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

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

\(=9-\frac{9}{10}=\frac{81}{10}\)