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2 tháng 3 2017

1) 

A= \(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{39.40}\)

\(\Rightarrow A=\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-...+\frac{1}{39}-\frac{1}{40}\)

\(\Rightarrow A=\frac{1}{3}-\frac{1}{40}\)

=> A= 27/120

2 tháng 3 2017

A = \(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{39.40}\)

\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{39}-\frac{1}{40}\)

\(\frac{1}{3}-\frac{1}{40}\)

\(\frac{37}{120}\)

B = \(\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{37.40}\)

\(\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{37}-\frac{1}{40}\right)\)

\(\frac{1}{3}\left(\frac{1}{4}-\frac{1}{40}\right)\)

\(\frac{1}{3}.\frac{9}{40}=\frac{3}{40}\)

C = \(\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{37.40}\)

\(\frac{2}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{37}-\frac{1}{40}\right)\)

\(\frac{2}{3}.\left(\frac{1}{4}-\frac{1}{40}\right)\)

\(\frac{2}{3}.\frac{9}{40}=\frac{3}{20}\)

NM
10 tháng 3 2022

ta nhân 3 cả hai vế, được : 

\(\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{102.105}\right)x=3\)

hay 

\(\left(\frac{4-1}{1.3}+\frac{7-4}{4.7}+...+\frac{105-102}{102.105}\right)x=3\) \(\Leftrightarrow\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+..+\frac{1}{102}-\frac{1}{105}\right)x=3\)

\(\Leftrightarrow\left(1-\frac{1}{105}\right)x=3\Leftrightarrow\frac{104}{105}.x=3\Leftrightarrow x=\frac{315}{104}\)

15 tháng 5 2017

\(3B=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}.\)

\(3B=\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+...+\frac{103-100}{100.103}\)

\(3B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}=1-\frac{1}{103}=\frac{102}{103}\)

\(B=\frac{102}{3.103}=\frac{34}{103}\)

15 tháng 5 2017
3B=3/1.4+3/4.7+3/7.10+...+3/100.103 3B=(4-1)/1.4+(7-4)/4.7+(10-7)/7.10+...+(103-100)/100.103 3B=1-1/4+1/4-1/7+1/7-1/10+...+1/100-1/103=1-1/103=102/103 B=102/(3.103)=34/103
28 tháng 8 2016

đề sai rồi sao lại có +100+103

9 tháng 4 2017

a)\(P=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{46}-\frac{1}{56}\)

=\(1-\frac{1}{56}=\frac{55}{56}\)

b)\(A.\frac{1}{3}=\frac{1}{3}.\left(\frac{3}{1.2}+\frac{3}{2.3}+....+\frac{3}{99.100}\right)\)

\(\frac{1}{3}A=\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{3}{99.100}\)

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

=> \(\frac{1}{3}A=1-\frac{1}{100}=\frac{99}{100}\)

=> \(A=\frac{99}{100}.3=\frac{297}{100}\)

c)\(B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\)

=\(1-\frac{1}{103}=\frac{102}{103}\)

d) \(\frac{3}{5}C=\frac{3}{5}.\left(\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\right)\)

=\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\)

=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{100}-\frac{1}{103}\)

=\(1-\frac{1}{103}=\frac{102}{103}\)

=>\(C=\frac{102}{103}.\frac{5}{3}=\frac{170}{103}\)

e) \(\frac{4}{7}D=\frac{4}{7}.\left(\frac{7}{1.5}+\frac{7}{5.9}+...+\frac{7}{101.105}\right)\)

=\(\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{101.105}\)

=\(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{101}-\frac{1}{105}\)

=\(1-\frac{1}{105}=\frac{104}{105}\)

=< D=\(\frac{104}{105}.\frac{7}{4}=\frac{26}{15}\)

17 tháng 7 2018

Đặt biểu thức trên là A. Ta có:

3A = 3/1.4 + 3/4.7 + 3/7.10 + ... + 3/2016/2019

3A = 1-1/4 +1/4-1/7+1/7-1/10/+ ... + 1/2016-1/2019

3A = 1-1/2019=2018/2019

A =1009/2019

17 tháng 7 2018

Ta có:

\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{2016.2019}\)

\(=\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{2016.2019}\right)\)

\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{2016}-\frac{1}{2019}\right)\)

\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{2019}\right)\)

\(=\frac{1}{3}.\frac{2018}{2019}\)

\(=\frac{2018}{6057}\)

3 tháng 5 2020

\(S_1=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{48\cdot49}+\frac{1}{49\cdot50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{48}-\frac{1}{49}+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}=\frac{49}{50}\)

\(S_2=\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\frac{1}{10\cdot13}+....+\frac{1}{94\cdot97}+\frac{1}{97\cdot100}\)

\(3S_2=\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+....+\frac{3}{94\cdot97}+\frac{3}{97\cdot100}\)

\(=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+....+\frac{1}{97}-\frac{1}{100}\)

\(=\frac{1}{4}-\frac{1}{100}=\frac{6}{25}\)

=> \(S_2=\frac{6}{25}:3=\frac{2}{25}\)

7 tháng 6 2021

\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+.......+\dfrac{1}{x\cdot\left(x+1\right)}=\dfrac{122}{123}\)

\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+......+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{122}{123}\)

\(\Leftrightarrow1-\dfrac{1}{x+1}=\dfrac{122}{123}\)

\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{123}\)

\(\Leftrightarrow x=122\)

7 tháng 6 2021

đây là toán lớp 5 mà có cả kí hiệu toán lớp 8 rồi giỏi ghê

 

19 tháng 3 2017

\(A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2015.2017}\)

\(A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\)

\(A=1-\dfrac{1}{2017}=\dfrac{2016}{2017}\)

\(B=\dfrac{3}{1.4}+\dfrac{3}{5.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}\)

\(B=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\)

\(B=1-\dfrac{1}{103}=\dfrac{102}{103}\)

\(C=\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{62.65}\)

\(3C=3\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{62.65}\right)\)

\(3C=\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{62.65}\)

\(3C=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{62}-\dfrac{1}{65}\)

\(3C=\dfrac{1}{2}-\dfrac{1}{65}\)

\(3C=\dfrac{63}{130}\)

\(C=\dfrac{63}{130}:3=\dfrac{21}{130}\)

19 tháng 3 2017

thanhk

haha

7 tháng 6 2018

làm dài lắm,nếu muốn thì k minh còn ko thì thôi

7 tháng 6 2018

a,0,36.350+1,2.20.3+9.4.4,5

=13.3.35+12.2.3+9.2.3.3

=3.(13.35+12.2+.9.2.3)

=3.(455+24+54)

=3.533

=1599

b,2015.2016-5/2015.2015+2010

=4062240-5+2010

=4064245

c,2/1.3+2/3.5+2/5.7+...+2/71.73

=1-1/3+1/3-1/5+1/5-1/7+...+1/71-1/73

=1-1/73

=72/73

d,(1+1/2).(1+1/3)+...+(1+1/2018)

=3/2.4/3.5/4+...+2019/2018

=2019/2

e,E=1/4.5+1/5.6+1/6.7+...+1/80.81(làm tương tự với phần d nên mình làm ngắn

     =1/4-1/81

     =77/324

f,F=3/2.3+3/3.4+...+3/99.100

=3.(1/2.3+1/3.4+...+1/99.100)(làm tương tự với d

=3.(1/2-1/100)

=3.49/100

=147/100

gG=5/1.4+5/4.7+...+5/61.64

3G=5.(3/1.4+3./4.7+...+3/61.64)

     =5.(1-1/64)

     =5.63/64

     =315/64

ok nha bạn,mình giữ đúng lời hứa.