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

B= \(-\frac{9}{10}=-0,9\)

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

\(=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{9x10}\)

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

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

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

\(\frac{1}{1.2}\)\(+\frac{1}{2.3}\)\(+\frac{1}{3.4}\)\(+...+\frac{1}{9.10}\)
\(1-\frac{1}{2}\)\(+\frac{1}{2}\)\(-\frac{1}{3}\)\(+\frac{1}{3}\)\(-\frac{1}{4}\)\(+...+\frac{1}{9}\)\(-\frac{1}{10}\)

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

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\)=\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{9x10}\)

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

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

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

a; [6.(- \(\dfrac{1}{3}\))³ - 3.(- \(\dfrac{1}{3}\) + 1)] - ( - \(\dfrac{1}{3}\) - 1)

=  [6. \(\dfrac{-1}{3^3}\) - 3.\(\dfrac{2}{3}\)] -  ( - \(\dfrac{1}{3}\) - \(\dfrac{3}{3}\)) 

= [\(\dfrac{-2}{9}\) - 2] + \(\dfrac{4}{3}\)

= [\(\dfrac{-2}{9}\) - \(\dfrac{18}{9}\)] + \(\dfrac{12}{9}\)

=   - \(\dfrac{20}{9}\) + \(\dfrac{12}{9}\)

=    \(\dfrac{-8}{9}\)

b; (6³ + 3.6² + 3³): 13

=  (216 + 3.36 + 27) : 13

= (216 + 108 + 27): 13

= (324 + 27): 13

= 351 : 13

= 27 

1/2+1/6+1/12+...+1/90=1/(1.2)+1/(2.3)+1/(3.4)+...+1/(9.10)

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

                                =1-1/10

                                =9/10

9/10 nha bạn

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+..+\frac{1}{90}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

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

dấu "." là nhân nhé

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

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

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

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

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

1/2+1/6+1/12+1/20+...+1/90

=1/1.2+1/2.3+1/3.4+...+1/9.10

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

=1-1/10

=9/10

\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}...+\frac{1}{9.10}\)

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