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b.ta chia B thành 10 nhóm mỗi nhóm có 6 hạng tử  \(B=\left(2+2^2+2^3+2^4+2^5+2^6\right)+....+\left(2^{55}+2^{56}+2^{57}+2^{58}+2^{59}+2^{60}\right)\)

\(B\text{=}2\left(1+2+2^2+2^3+2^4+2^5\right)+...+2^{55}\left(1+2+2^2+2^3+2^4+2^5\right)\)

\(B\text{=}2.63+...+2^{56}.63\)

\(\Rightarrow B⋮63\)

\(\Rightarrow B⋮21\)

bài 2

Bài 3

\(\dfrac{55}{23}+\dfrac{-22}{23}\le x\le\dfrac{1}{5}-\dfrac{-1}{6}+\dfrac{79}{30}\)

\(=\dfrac{33}{23}\)\(\le x\le\dfrac{90}{30}\)

\(=\dfrac{33}{23}\le x\le3\)

Mà \(x\in Z\) \(\Rightarrow\)\(x=2\)

Có 1 giá trị thỏa mãn 

Chọn A

Bài 4

\(\dfrac{-11}{12}< \dfrac{5}{x}< \dfrac{-11}{15}\)

Chọn D

Bài 5

\(M=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)

\(M=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\)

\(M=1-\dfrac{1}{100}\)

\(M=\dfrac{100}{100}-\dfrac{1}{100}\)

\(M=\dfrac{99}{100}\)

CHọn C

A
D
C

 cả dãy đang trừ mà sao cái cuối là cộng vậy bạn, dãy ko có quy tắc à :v

Hmm đề sai :v 

Sửa lại:

`1+1/(1-1/2)+1/(1-/3)+1/(1-3/4)+...........+1/(1+n/(n+1))=276`

`=>1+1/(1/2)+1/(1/3)+1/(1/4)+......+1/(1/n)=276`

`=>1+2+3+4+.......+n=276`

Từ `1->n` có n số 
`=>1+2+3+4+.....+n=(n(n+1))/2`
`=>(n(n+1))/2=276`
`=>n(n+1)=552`
`=>n(n+1)=23.24`
`=>n=23`
Vậy `n=23`

f)\(-\dfrac{2}{17}+\dfrac{15}{23}+-\dfrac{15}{17}+\dfrac{4}{19}+\dfrac{8}{23}\)
\(=\left(-\dfrac{2}{17}+-\dfrac{15}{17}\right)+\left(\dfrac{15}{23}+\dfrac{8}{23}\right)+\dfrac{4}{19}\)
\(=-1+1+\dfrac{4}{19}\)
\(=0+\dfrac{4}{19}=\dfrac{4}{19}\)
g)\(-\dfrac{1}{2}+\dfrac{3}{21}+-\dfrac{2}{6}+-\dfrac{5}{30}\)
\(=-\dfrac{1}{2}+\dfrac{1}{7}+-\dfrac{1}{3}+-\dfrac{1}{6}\)
\(=\left(-\dfrac{1}{2}+-\dfrac{1}{3}+-\dfrac{1}{6}\right)+\dfrac{1}{7}\)
\(=-\dfrac{3+2+1}{6}+\dfrac{1}{7}\)
\(=\dfrac{1}{7}-1\)
\(=\dfrac{1}{7}-\dfrac{7}{7}=-\dfrac{6}{7}\)

`f)(-2)/17 + 15/23 + (-15)/17 + 4/19 + 8/23`

`= (-2/17+ -15/17)+(15/23+8/23)+4/19`

`= -1+1+4/19`

`= 0 +4/19`

`= 0`

`g)(-1)/2 + 3/21 + (-2)/6 + (-5)/30`

`= (-1)/2 + 1/7 + (-1)/3 + (-1)/6`

`= (-21)/42 + 6/42 + (-14)/42 + (-7)/42`

`=(-36)/42`

`=(-6)/7`

\(f,-\dfrac{2}{17}+\dfrac{15}{23}-\dfrac{15}{17}+\dfrac{4}{19}+\dfrac{8}{23}\)

\(=-\dfrac{2}{17}-\dfrac{15}{17}+\dfrac{15}{23}+\dfrac{8}{23}+\dfrac{4}{19}\)

\(=-1+1+\dfrac{4}{19}\)

\(=\dfrac{4}{19}\)

\(g,-\dfrac{1}{2}+\dfrac{3}{21}-\dfrac{2}{6}-\dfrac{5}{30}\)

\(=-\dfrac{5}{14}-\dfrac{1}{2}\)

\(=-\dfrac{5}{14}-\dfrac{7}{14}\)

\(=-\dfrac{6}{7}\)

a) Ta có: \(\dfrac{-3}{7}+\dfrac{15}{26}-\left(\dfrac{2}{13}-\dfrac{3}{7}\right)\)

\(=\dfrac{-3}{7}+\dfrac{15}{26}-\dfrac{2}{13}+\dfrac{3}{7}\)

\(=\dfrac{15}{26}-\dfrac{4}{26}\)

\(=\dfrac{11}{26}\)

b) Ta có: \(2\cdot\dfrac{3}{7}+\left(\dfrac{2}{9}-1\dfrac{3}{7}\right)-\dfrac{5}{3}:\dfrac{1}{9}\)

\(=\dfrac{6}{7}+\dfrac{2}{9}-\dfrac{10}{7}-\dfrac{5}{3}\cdot9\)

\(=\dfrac{-4}{7}+\dfrac{2}{9}-15\)

\(=\dfrac{-36}{63}+\dfrac{14}{63}-\dfrac{945}{63}\)

\(=\dfrac{-967}{63}\)

c) Ta có: \(\dfrac{-11}{23}\cdot\dfrac{6}{7}+\dfrac{8}{7}\cdot\dfrac{-11}{23}-\dfrac{1}{23}\)

\(=\dfrac{-11}{23}\cdot\left(\dfrac{6}{7}+\dfrac{8}{7}\right)-\dfrac{1}{23}\)

\(=\dfrac{-11}{23}\cdot2-\dfrac{1}{23}\)

\(=-1\)

d) Ta có: \(\left(\dfrac{377}{-231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot\left(\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{24}\right)\)

\(=\left(\dfrac{-377}{231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot\left(\dfrac{4}{24}-\dfrac{3}{24}-\dfrac{1}{24}\right)\)

\(=\left(\dfrac{-377}{231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot0\)

=0