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Ta có:A = 1/1.2 + 1/2.3 + 1/3.4 + ...+ 1/2020.2021

         A=1-1/2+1/2-1/3+1/3-1/4+...+1/2020-1/2021

        A=1-1/2021

Ta có: B = 1/6 + 1/12 + 1/20 + ... + 1/240

          B=1/2.3+1/3.4+1/4.5+....+1/15.16

           B=1/2-1/3+1/3-1/4+1/4-1/5+....+1/15-1/16

          B=1/2-1/16

phần C bn có đánh nhầm đề bài ko

15 tháng 6 2020

\(a)\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{132}\)

\(=\frac{22}{132}+\frac{11}{132}+\frac{1}{20}+\frac{1}{132}\)

\(=\frac{33}{132}+\frac{1}{20}+\frac{1}{132}\)

\(=\frac{34}{132}+\frac{1}{20}\)

\(=\frac{17}{66}+\frac{1}{20}\)

\(=\frac{203}{660}\)

15 tháng 6 2020

\(a,\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{132}\) 

\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{132}\)

\(=\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\right)+\frac{1}{132}\)

\(=\left(\frac{1}{2}-\frac{1}{5}\right)+\frac{1}{132}\)

\(=\frac{3}{10}+\frac{1}{132}\)

\(=\frac{198}{660}+\frac{5}{660}\)

\(=\frac{203}{660}\)

16 tháng 3 2018

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

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

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

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

Ta có \(B=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)

\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)

\(=\frac{1}{2}-\frac{1}{7}\)

\(=\frac{5}{14}\)

Ta có \(C=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)

\(=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)

\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{11}\right)\)

\(=\frac{1}{6}-\frac{1}{22}\)

\(=\frac{4}{33}\)

16 tháng 3 2018

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=1-\frac{1}{100}\)

\(A=\frac{99}{100}\)

\(B=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)

\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)

\(B=\frac{1}{2}-\frac{1}{7}\)

\(B=\frac{5}{14}\)

\(C=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)

\(C=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)

\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}\right)\)

\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{11}\right)\)

\(C=\frac{1}{6}-\frac{1}{22}=\frac{4}{33}\)

4 tháng 11 2016

A=\(\frac{19}{20}\),

4 tháng 11 2016

sao ra vay ban minh muoc cach giai bai

29 tháng 3 2022

\(1-\dfrac{1}{x+1}=\dfrac{14}{15}\)

\(\dfrac{x+1-1}{x+1}=\dfrac{14}{15}\)

\(\dfrac{x}{x+1}=\dfrac{14}{15}\)

\(15x=14x+14\)

\(x=14\)

3 tháng 5 2022

\(\text{#}HaimeeOkk\)

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2018.2019}+\dfrac{1}{2019.2020}\)

\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2018}-\dfrac{1}{2019}+\dfrac{1}{2019}-\dfrac{1}{2020}\)

\(A=1-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-\left(\dfrac{1}{4}-\dfrac{1}{4}\right)-...-\left(\dfrac{1}{2019}-\dfrac{1}{2019}\right)-\dfrac{1}{2020}\)

\(A=1-0-0-0-...-0-\dfrac{1}{2020}\)

\(A=1-\dfrac{1}{2020}\)

\(A=\dfrac{2019}{2020}\)

Vậy \(A=\dfrac{2019}{2020}\)

6 tháng 5 2017

1.

a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{98}{99}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x-1}=\frac{98}{99}\)

\(1-\frac{1}{x-1}=\frac{98}{99}\)

\(\frac{1}{x-1}=1-\frac{98}{99}\)

\(\frac{1}{x-1}=\frac{1}{99}\)

\(\Rightarrow x-1=99\)

\(\Rightarrow x=99+1=100\)

b) \(x-\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-...-\frac{20}{53.55}=\frac{3}{11}\)

\(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)

\(x-\left[10.\left(\frac{1}{11}-\frac{1}{13}\right)+10.\left(\frac{1}{13}-\frac{1}{15}\right)+10.\left(\frac{1}{15}-\frac{1}{17}\right)+...+10.\left(\frac{1}{53}-\frac{1}{55}\right)\right]=\frac{3}{11}\)

\(x-\left[10.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)\right]=\frac{3}{11}\)

\(x-\left[10.\left(\frac{1}{11}-\frac{1}{55}\right)\right]=\frac{3}{11}\)

\(x-10.\frac{4}{55}=\frac{3}{11}\)

\(x-\frac{8}{11}=\frac{3}{11}\)

\(\Rightarrow x=\frac{3}{11}+\frac{8}{11}=1\)

c) 5x + 2 . 5x + 23 = 83

5x . ( 1 + 2 ) + 8 = 83

5x . 3 = 83 - 8

5x . 3 = 75

5x = 75 : 3

5x = 25

\(\Rightarrow\)5x = 52

\(\Rightarrow\)x = 2

2.

Ta thấy \(2016^{2016}>2016^{2016}-3\)

\(\Rightarrow B=\frac{2016^{2016}}{2016^{2016}-3}>\frac{2016^{2016}+2}{2016^{2016}-3+2}=\frac{2016^{2016}+2}{2016^{2016}-1}=A\)

\(\Rightarrow A< B\)

6 tháng 5 2017

a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{98}{99}\)

Ta có \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{98}{99}\)

      = \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{98}{99}\)(áp dụng công thức)

      = \(1-\frac{1}{x+1}=\frac{98}{99}\)

      = \(\frac{1}{x+1}=1-\frac{98}{99}\)(quy tắc tìm số trừ)

      = \(\frac{1}{x+1}=\frac{1}{99}\Rightarrow\frac{1}{x+1}=\frac{1}{98+1}\Rightarrow x=98\)

Vậy x = 98 :)

Còn nữa, công thức mà mình áp dụng là: \(\frac{a}{b.c}=\frac{1}{b}-\frac{1}{c}\)nếu \(a=c-b\)

8 tháng 3 2016

T-i-ck nha,k nha, câu trả lờii sẽ hiện ra

8 tháng 3 2016

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

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

\(B=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)

\(B=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)

\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)

11 tháng 7 2016

\(A=\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{49.50}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=\frac{1}{1}-\frac{1}{50}\)

\(A=\frac{49}{50}\)

Vì \(\frac{245}{420}< \frac{245}{294}< \frac{245}{250}\)

Vậy \(\frac{7}{12}< \frac{49}{50}< \frac{5}{6}\)