Đặt A=\(\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{60}\)

A=\(\left(\dfrac{20}{20.21}+\dfrac{21}{21.22}+...+\dfrac{39}{39.40}\right)+\left(\dfrac{40}{40.41}+\dfrac{41}{41.42}+...+\dfrac{59}{59.60}\right)\)

=>A >\(20\cdot\left(\dfrac{1}{20.21}+\dfrac{1}{21.22}+...+\dfrac{1}{39.40}\right)+40\cdot\left(\dfrac{1}{40.41}+\dfrac{1}{41.42}+...+\dfrac{1}{59.60}\right)\)

A>\(20\cdot\left(\dfrac{1}{20}-\dfrac{1}{40}\right)+40\cdot\left(\dfrac{1}{40}-\dfrac{1}{60}\right)=\dfrac{5}{6}>\dfrac{11}{15}\)

Mặt khác: A<\(40\cdot\left(\dfrac{1}{20.21}+\dfrac{1}{21.22}+...+\dfrac{1}{39.40}\right)+60\cdot\left(\dfrac{1}{40.41}+\dfrac{1}{41.42}+...+\dfrac{1}{59.60}\right)\)

A<\(40\cdot\left(\dfrac{1}{20}-\dfrac{1}{40}\right)+60\cdot\left(\dfrac{1}{40}-\dfrac{1}{60}\right)=\dfrac{3}{2}\)

Vậy...

bài 2

\(a.\)

\(\dfrac{17}{8}:\left(\dfrac{27}{8}+\dfrac{11}{2}\right)\)

\(=\dfrac{17}{8}:\left(\dfrac{27+44}{8}\right)=\dfrac{17}{8}:\dfrac{71}{8}=\dfrac{17}{8}\cdot\dfrac{8}{71}=\dfrac{17}{71}\)

\(b.\)

\(\dfrac{28}{15}\cdot\dfrac{1}{4^2}\cdot3+\left(\dfrac{8}{15}-\dfrac{69}{60}\cdot\dfrac{5}{23}\right):\dfrac{51}{54}\)

\(=\dfrac{28}{15}\cdot\dfrac{1}{4^2}\cdot3+\left(\dfrac{8}{15}-\dfrac{1}{4}\right):\dfrac{51}{54}\)

\(=\dfrac{28}{15}\cdot\dfrac{1}{4^2}\cdot3+\left(\dfrac{8\cdot4-15}{60}\right):\dfrac{51}{54}\)

\(=\dfrac{28}{15}\cdot\dfrac{1}{4^2}\cdot3+\dfrac{17}{60}:\dfrac{51}{54}\)

\(=\dfrac{28}{15}\cdot\dfrac{1}{16}\cdot3+\dfrac{17}{60}\cdot\dfrac{54}{51}\)

\(=\dfrac{7}{20}+\dfrac{3}{10}\)

\(=\dfrac{7+3\cdot2}{20}=\dfrac{13}{20}\)

Sai kìa phải bằng 11/20

 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`

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

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

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

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

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

\(\Rightarrow B⋮63\)

\(\Rightarrow B⋮21\)