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

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

\(=\frac{1}{90}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)\)

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

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

\(=\frac{-79}{90}\)

1/90 - 1/72 - 1/56 - ... - 1/6 - 1/2

= 1/90 - (1/2 + 1/6 + ... + 1/56 + 1/72)

= 1/90 - (1/1×2 + 1/2×3 + ... + 1/7×8 + 1/8×9)

= 1/90 - (1 - 1/2 + 1/2 - 1/3 + ... + 1/7 - 1/8 + 1/8 - 1/9)

= 1/90 - (1 - 1/9)

= 1/90 - 8/9

= 1/90 - 80/90

= -79/90

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

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

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

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

\(=\frac{-79}{90}\)

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

A=1/20+1/30+1/42+1/56+1/72+1/90

A=1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10

A=1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10

A=1/4-1/10

A=3/20

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

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

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

\(=\frac{1}{2}-\frac{1}{10}\)

\(=\frac{4}{10}=\frac{2}{5}\)

Ủng hộ mk nha !!! ^_^

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

\(=\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+....+\frac{1}{9x10}\)

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

\(=\frac{1}{2}-\frac{1}{10}=\frac{5}{10}-\frac{1}{10}=\frac{4}{10}=\frac{2}{5}\)

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

=\(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{9\times10}\) 

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}\)

\(=\frac{1}{2}-\frac{1}{10}\)

\(=\frac{5}{10}-\frac{1}{10}\)

\(\frac{2}{5}\)

1/90 - 1/72 - 1/56 - 1/42 - 1/30 - 1/20 - 1/12 - 1/6 - 1/2

= 1/90 - ( 1/72 + 1/56 + 1/42 + 1/30 + 1/20 + 1/12 + 1/6 + 1/2)

= 1/90 - ( 1/2 + 1/6 + 1/12 + ...+ 1/72)

= 1/90 - ( 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/8.9)

= 1/90 - ( 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/8 - 1/9)

= 1/90 - ( 1 - 1/9)

= 1/90 - 8/9

= 1/90 - 80/90

= -79/90

1/90 - 1/72 - 1/56 - 1/42 - 1/30 - 1/20 - 1/12 - 1/6 - 1/2

= 1/90 - ( 1/72 + 1/56 + 1/42 + 1/30 + 1/20 + 1/12 + 1/6 + 1/2)

= 1/90 - ( 1/2 + 1/6 + 1/12 + ...+ 1/72)

= 1/90 - ( 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/8.9)

= 1/90 - ( 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/8 - 1/9)

= 1/90 - ( 1 - 1/9)

= 1/90 - 8/9

= 1/90 - 80/90

= -79/90

 mk nha cac ban 

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

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

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

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 \(\frac{9}{10}\)

bằng 10/11

đúng nhớ **** cho minh nha

1 + 1 = 11 / 2 + 2 = 22

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T-T

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

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

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

=1-1/10 

=9/10

Me too ! 

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}=x-\left(2+4+..+100\right)\)

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

         \(\left(2+4+...+100\right)\)là \(B\). Ta có :  

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

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

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

Số số hạng của \(B\)là: \(\left(100-2\right)\div2+1=50\)

Tổng của \(B\)là: \(\left(2+100\right)\times50\div2=2550\)

\(\Rightarrow\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}=x-\left(2+4+..+100\right)=\frac{9}{10}=x-2550\)

\(\Rightarrow x=2550+\frac{9}{10}=2550+0,9=2550,9\)