89/90 Đấy

89/90

T=1/2+1/6+1/12+1/20+1/30+...+1/72+1/90

=> T=1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+...+1/9.10

=> T=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+...+1/9-1/10

=> T=1/1-1/10

=> T=9/10

em lớp 6  nha

B= 1/2 + 1/6 + 1/12 +1/20 + 1/30 + 1/42 + 1/56 + 1/72

B= 1/1*2 + 1/2*3 + 1/3*4 + 1/4*5 + 1/5*6 + 1/6*7 + 1/7*8 + 1/8*9

B=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9

B=1+0-0-0-0-0-0-0-1/9

B=1-1/9

B=8/9

k em nha

phần A em chịu

còn câu dưới đó e

=-79/90

\(\text{Ta có: }\)\(\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)

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

\(=\frac{1}{90}-\left(\frac{1}{9.8}+\frac{1}{8.7}+\frac{1}{7.6}+\frac{1}{6.5}+\frac{1}{5.4}+\frac{1}{4.3}+\frac{1}{3.2}+\frac{1}{2.1}\right)\)

\(=\frac{1}{90}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{8}-\frac{1}{9}\right)\)

\(=\frac{1}{90}-\left(1-\frac{1}{9}\right)\)

\(=\frac{1}{90}-\frac{8}{9}=-\frac{81}{90}=-\frac{9}{10}\)

<=> 

D = 1/90+1/72+1/56+1/42+1/30+1/20+1/12+1/6+1/2

D = 1/(1x2) + 1/(2x3) + 1/(3x4) + 1/(4x5) + 1/(5x6) + … + 1/(9x10)

D = 1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 + …. + 1/9 – 1/10

D = 1 – 1/10

D = 9/10

=1/(9.10)-1/(8.9)-1/(7.8)-1/(6.7)-1/(5.6)-1/(4.5)-1/(3.4)-1/(2.3)-1/(1.2)

sau đó làm tiếp nha bạn

( 1/90 + 1/72 + ... + 1/2)
= - { 1/( 9.10) + 1/(9.8) + ... + 1/( 2.1)}
= - ( 1/9 - 1/10 + 1/8 - 1/9 + ...+ 1 - 1/2) ( 1/90 = 1/(9.10) = 1/9 - 1/10)
= - ( 1 - 1/10)
= -9/10

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

\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+.....+\frac{1}{9\cdot10}\)

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

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

\(\frac{1}{2}+\frac{1}{6}+.......+\frac{1}{90}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{9.10}\)

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

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

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

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

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

\(=>?=\frac{1}{10\cdot11}=\frac{1}{110}\)

Vậy \(?\) là \(\frac{1}{110}\)

= 0,9 nhé

CHÚC BẠN HỌC GIỎI

TK MÌNH NHÉ

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

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

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

\(=>?=1-\frac{1}{10}=\frac{9}{10}\)

Vậy kết quả phép tính trên là \(\frac{9}{10}\)