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24 tháng 4 2015

a)\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}\)

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

\(=\left(1-\frac{1}{6}\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+...+\left(\frac{1}{5}-\frac{1}{5}\right)\)

\(=\left(1-\frac{1}{6}\right)+0+...+0=1-\frac{1}{6}=\frac{6}{6}-\frac{1}{6}=\frac{5}{6}\)

b)\(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\)

\(=\left(\frac{1}{2}-\frac{1}{14}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{8}-\frac{1}{8}\right)+\left(\frac{1}{11}-\frac{1}{11}\right)\)

\(=\left(\frac{1}{2}-\frac{1}{14}\right)+0+...+0=\frac{1}{2}-\frac{1}{14}=\frac{7}{14}-\frac{1}{14}=\frac{6}{14}\)

Nhớ **** cho mình nhé bạn! chúc bạn học tốt

24 tháng 4 2015

tick đúng cho mik nha please

Giải:

A=1/10+1/40+1/88+1/154+1/238+1/340

A=1/2.5+1/5.8+1/8.11+1/11.14+1/14.17+1/17.20

A=1/2-1/5+1/5-1/8+1/8-1/11+1/11-1/14+1/14-1/17+1/17-1/20

A=1/2-1/20

A=9/20

D=1/3+1/6+1/12+1/24+1/48

D=1/3+1/2.3+1/3.4+1/4.6+1/6.8

D=1/3+1/2-1/3+1/3-1/4+1/2.(2/4.6+2/6.8)

D=1/3+1/2-1/4+1/2.(1/4-1/6+1/6-1/8)

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

D=1/3+1/4+1/2.1/8

D=1/3+1/4+1/16

D=31/48

F=0,5-1/3-0,4-4/7-1/6+4/35-1/41

F=1/2-1/3-2/5-4/7-1/6+4/35-1/41

F=1/6-(-6/35)-1/6+4/35-1/41

F=(1/6-1/6)+(6/35+4/35)-1/41

F=0+2/7-1/41

F=2/7+1/41

F=75/287

Chúc bạn học tốt!

24 tháng 4 2021

Cảm ơn nhìu <3

7 tháng 3 2022

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7 tháng 3 2022

Mình cảm ơn nha

 

mk làm phần a thui nhé

a. A = 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6

A = 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6

A =  1/2 - 1/6

A= 3/6 - 1/6

A = 1/3

10 tháng 11 2019

\(B=\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}\)

\(b=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\)

\(b=\frac{1}{2}-\frac{1}{14}\)

\(b=\frac{3}{7}\)

\(d=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\)

\(d=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{10\cdot11}\)

\(d=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\)

\(d=1-\frac{1}{11}\)

\(d=\frac{10}{11}\)

\(e=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)

\(e=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+\frac{1}{14\cdot17}+\frac{1}{17\cdot20}\)

\(e=\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{17\cdot20}\right)\)

\(e=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\right)\)

\(e=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(e=\frac{1}{3}\cdot\frac{9}{20}=\frac{3}{20}\)

\(\dfrac{1}{2}\cdot1\cdot\dfrac{1}{3}\cdot10\cdot\dfrac{7}{35}\cdot\dfrac{3}{4}\)

\(=\dfrac{1}{2}\cdot10\cdot\dfrac{1}{5}\cdot1\cdot\dfrac{1}{3}\cdot\dfrac{3}{4}\)

\(=\dfrac{10}{10}\cdot\dfrac{1}{4}=\dfrac{1}{4}\)

Bài 1: 

a: \(\dfrac{25}{42}-\dfrac{20}{63}=\dfrac{75-40}{126}=\dfrac{35}{126}=\dfrac{5}{18}\)

b: \(\dfrac{9}{20}-\dfrac{13}{75}-\dfrac{1}{6}=\dfrac{135}{300}-\dfrac{52}{300}-\dfrac{50}{300}=\dfrac{33}{300}=\dfrac{11}{100}\)

a: =382-282+531-331

=100+200=300

b: =(7-8)+(9-10)+...+(2009-2010)

=(-1)+(-1)+....+(-1)

=-1002

c: =-(1+2+3+...+2009+2010)

=-2010*2011/2=-2021055

10 tháng 2 2023

`1 - 2 + 3 - 4 + 5 - 6 +...+ 2021 - 2022`

`= (1 - 2) + (3 - 4) + (5 - 6) +...+ (2021 - 2022)`

`= (-1) + (-1) + (-1) + ... + (-1) `  [có `2022 : 2 = 1011` nhóm]

`= (-1) xx 1011 = -1011`