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26 tháng 6 2016

\(A=1+1+1+1-\left(\frac{1}{3}-\frac{1}{15}-\frac{1}{35}-\frac{1}{63}\right)\)

\(A=4+\left(\frac{1}{1.3}-\frac{1}{3.5}-\frac{1}{5.7}-\frac{1}{7.9}\right)\)

\(A=4+\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)\)

\(A=4+\left(1-\frac{1}{9}\right)\)

\(A=4+\frac{8}{9}=\frac{44}{9}\)

Vậy A=44/9

7 tháng 7 2019

\(3x-\frac{1}{3}-\frac{1}{15}-\frac{1}{35}-\frac{1}{63}-\frac{1}{99}=0\)

\(\Rightarrow3x-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)=0\)

\(\Rightarrow3x-\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)=0\)

\(\Rightarrow3x-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)=0\)

\(\Rightarrow3x-\left(1-\frac{1}{99}\right)=0\)

\(\Rightarrow3x-\frac{98}{99}=0\)

\(\Rightarrow3x=0+\frac{98}{99}\)

\(\Rightarrow3x=\frac{98}{99}\)

\(\Rightarrow x=\frac{98}{99}:3\)

\(\Rightarrow x=\frac{98}{297}\)

7 tháng 7 2019

\(3x-\frac{1}{3}-\frac{1}{15}-\frac{1}{35}-\frac{1}{63}-\frac{1}{99}=0\)

\(2\left(3x-\frac{1}{3}-\frac{1}{15}-\frac{1}{35}-\frac{1}{63}-\frac{1}{99}\right)=2.0\)

\(6x-\frac{2}{3}-\frac{2}{15}-\frac{2}{35}-\frac{2}{63}-\frac{2}{99}=0\)

\(6x-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)=0\)

\(6x-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)=0\)

\(6x-\left(1-\frac{1}{11}\right)=0\)

\(6x-\frac{10}{11}=0\)

\(6x=\frac{10}{11}\)

\(x=\frac{5}{33}\)

26 tháng 9 2019

A = \(\frac{1}{3}+\frac{13}{35}+\frac{33}{35}+\frac{61}{63}+\frac{97}{99}+\frac{141}{143}\)

\(=\left(1-\frac{2}{3}\right)+\left(1-\frac{2}{15}\right)+\left(1-\frac{2}{35}\right)+\left(1-\frac{2}{63}\right)+\left(1-\frac{2}{99}\right)+\left(1-\frac{2}{143}\right)\)

\(=\left(1+1+1+1+1+1\right)-\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\right)\)

\(=6-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)

\(=6-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)

\(=6-\left(1-\frac{1}{13}\right)\)

\(=6-1+\frac{1}{13}\)

\(=5+\frac{1}{13}\)

\(=\frac{66}{13}\)

\(\text{Vậy }A=\frac{66}{13}\)

6 tháng 7 2019

a)\(\frac{1}{2}-2.\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+.....+\frac{1}{48.50}\right)\)

=\(\frac{1}{2}-\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+.....+\frac{2}{48.50}\right)\)

=\(\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+.....+\frac{1}{48}-\frac{1}{50}\right)\)

=\(\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{50}\right)\)

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

6 tháng 7 2019

\(1)a)\frac{1}{2}-2\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{48.50}\right)\)

\(=\frac{1}{2}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{24.25}\right)\)

\(=\frac{1}{2}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{24}-\frac{1}{25}\right)\)

\(=\frac{1}{2}-\left(1-\frac{1}{25}\right)\)

\(=\frac{1}{2}-\frac{24}{25}=\frac{-23}{50}\)

\(\)

5 tháng 1 2016

\(A=\frac{-1}{3}+\frac{-1}{15}+\frac{-1}{35}+\frac{-1}{63}+...+\frac{-1}{9999}\)
\(A=-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\right)\)
\(\Rightarrow2A=-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99\cdot101}\right)\)
\(2A=-\left(2-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+\frac{2}{7}-\frac{2}{9}+...+\frac{2}{99}-\frac{2}{101}\right)\)
\(2A=-\left(2-\frac{2}{101}\right)\)
\(2A=-\frac{200}{101}\)
\(\Rightarrow A=-\frac{100}{101}\)

28 tháng 2 2018

Đặt biểu thức trên là A, ta có:

\(A=\frac{-1}{3}+\frac{-1}{15}+\frac{-1}{35}+\frac{-1}{63}+...+\frac{-1}{9999}\)

\(\Rightarrow A=-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\right)\)

\(\Rightarrow A=-\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\right)\)

\(\Rightarrow2A=-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\right)\)

\(\Rightarrow2A=-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(\Rightarrow2A=-\left(1-\frac{1}{101}\right)\)

\(\Rightarrow2A=-\frac{100}{101}\)

\(\Rightarrow A=-\frac{100}{101}\div2=-\frac{50}{101}\)

\(M=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right).....\left(\frac{1}{100}-1\right)\left(\frac{1}{121}-1\right)=\frac{-3}{4}.\frac{-8}{9}.....\frac{-99}{100}.\frac{-120}{121}\)

\(M=\frac{-1.3}{2.2}.\frac{-2.4}{3.3}.....\frac{-9.11}{10.10}.\frac{-10.12}{11.11}=\frac{-1}{2}.\frac{-12}{11}=\frac{12}{22}=\frac{6}{11}\)

\(S=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{99}\)

\(S=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{9.11}\right)\)

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

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

\(S=\frac{5}{11}\)

\(Q=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2013.2015}\)

\(Q=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)\)

\(Q=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)

\(Q=\frac{1}{2}\left(1-\frac{1}{2015}\right)\)

\(Q=\frac{1007}{2015}\)

~ Đấng Ed :) ~ 

17 tháng 8 2018

C = 1/3 + -3/4 + 3/5 + 1/57 + -1/36 + 1/15 + -2/9

C = ( 1/3 + 1/57 ) + ( -3/4 + -1/36 ) + ( 3/5 + 1/15 ) + -2/9 

C = ( 19/57 + 1/57 ) + ( -27/36 + -1/36 ) + ( 9/15 + 1/15 ) + -2/9 

C = 20/57 + -28/36 + 10/15 + -2/9 

C = 20/57 + -7/9 + 2/3 + -2/9

C = ( 20/57 + 2/3 ) + ( -7/9 + -2/9 )

C = 58/57 + -1 

C = 1/57

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

D = ( 1/2 + 1/3 + 1/6 ) + ( -1/5 + -5/7 +-3/35 ) + 1/41

D = ( 3/6 + 2/6 + 1/6 ) + ( -7/35 + -25/35 + -3/35 ) + 1/41

D = 1 + -1 + 1/41

D = 1/41

E = -1/2 + 3/5 + -1/9 + 1/127 + -7/18 + 4/35 + 2/7 

E = ( -1/2 + -1/9 + -7/18 ) + ( 3/5 + 4/35 ) + 1/127 + 2/7

E = ( -9/18 + -2/18 + -7/18 ) + ( 21/35 + 4/35 ) + 1/127 + 2/7

E = -1 + 5/7 + 1/257 + 2/7 

E = -1 + ( 5/7 + 2/7 ) + 1/127

E = -1 + 1 + 1/127

E = 1/127

17 tháng 8 2018

\(C=\frac{1}{3}+\frac{-3}{4}+\frac{3}{5}+\frac{1}{57}+\frac{-1}{36}+\frac{1}{15}+\frac{-2}{9}.\)

\(C=\left(\frac{1}{3}+\frac{3}{5}+\frac{1}{15}\right)-\left(\frac{3}{4}+\frac{1}{36}+\frac{2}{9}\right)+\frac{1}{57}\)

\(C=1-1+\frac{1}{57}\)

\(C=\frac{1}{57}\)

31 tháng 7 2015

Cho A chứng minh B, kì vậy ?

31 tháng 7 2015

Ta có: \(A=\frac{3-2}{3}+\frac{15-2}{15}+\frac{35-2}{35}+\frac{63-2}{63}+\frac{99-2}{99}+\frac{143-2}{143}+\frac{195-2}{195}\)

\(A=\left(1-\frac{2}{3}\right)+\left(1-\frac{2}{15}\right)+\left(1-\frac{2}{35}\right)+\left(1-\frac{2}{63}\right)+\left(1-\frac{2}{99}\right)+\left(1-\frac{2}{143}\right)+\left(1-\frac{2}{195}\right)\)

\(A=7-\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}+\frac{2}{195}\right)\)

\(A=7-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\right)\)

\(A=7-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)

\(A=7-\left(1-\frac{1}{15}\right)=7-1+\frac{1}{15}=6\frac{1}{15}\)không là số nguyên

30 tháng 5 2018

a) \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{100.103}\)

\(=\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}\right)\)

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

\(=\frac{1}{3}.\left(1-\frac{1}{103}\right)\)

\(=\frac{1}{3}.\frac{102}{103}\)

\(=\frac{34}{103}\)

b) \(\frac{1}{2000.1999}-\frac{1}{1999.1998}-\frac{1}{1998.1997}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(=\frac{1}{2000.1999}-\left(\frac{1}{1999.1998}+\frac{1}{1998.1997}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)(*)

Đặt biểu thức trong ngoặc là A ta có :

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1997.1998}+\frac{1}{1998.1999}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1997}-\frac{1}{1998}+\frac{1}{1998}-\frac{1}{1999}\)

\(A=1-\frac{1}{1999}\)

\(A=\frac{1998}{1999}\)

Thay vào biểu thức (*) ta có :

\(\frac{1}{2000.1999}-\frac{1998}{1999}\)

\(=\frac{1}{3998000}-\frac{1998}{1999}\)

\(=\frac{-3995999}{3998000}\)

c) \(\frac{-1}{3}+\frac{-1}{15}+\frac{-1}{35}+\frac{-1}{63}+...+\frac{-1}{9999}\)

\(=\frac{-1}{1.3}+\frac{-1}{3.5}+\frac{-1}{5.7}+\frac{-1}{7.9}+...+\frac{-1}{99.101}\)

\(=\frac{-1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\right)\)

\(=\frac{-1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{-1}{2}.\left(1-\frac{1}{101}\right)\)

\(=\frac{-1}{2}.\frac{100}{101}\)

\(=\frac{-50}{101}\)

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