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7 tháng 2 2020
Lời giải:
Xét tổng quát:
1+1k(k+2)=k(k+2)+1k(k+2)=(k+1)2k(k+2)1+1k(k+2)=k(k+2)+1k(k+2)=(k+1)2k(k+2)
Thay k=1,2,....,2015k=1,2,....,2015 ta có:
1+11.3=221.31+11.3=221.3
1+12.4=322.41+12.4=322.4
1+13.5=423.51+13.5=423.5
1+14.6=524.61+14.6=524.6
.............
1+12015.2017=201622015.20171+12015.2017=201622015.2017
Nhân theo vế:
⇒A=12(1+11.3)(1+12.4)(1+13.5)....(1+12015.2017)⇒A=12(1+11.3)(1+12.4)(1+13.5)....(1+12015.2017)
=12.221.3.322.4.423.5.524.6....201622015.2017=12.221.3.322.4.423.5.524.6....201622015.2017
=(1.2.3...2016)2(1.2.3...2015)(2.3.4...2017)=(1.2.3...2016)(2.3....2016)(1.2.3...2015)(2.3.4...2017)=2016.12017=20162017
Q
5
V
12 tháng 4 2017
=\(\frac{1}{2}x\left(\frac{2}{5x7}+\frac{2}{7x9}+\frac{2}{9x11}+...+\frac{2}{2015x2017}\right)\)
=\(\frac{1}{2}x\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
=\(\frac{1}{2}x\left(\frac{1}{5}-\frac{1}{2017}\right)\)
=\(\frac{1}{2}x\frac{2012}{10085}\)
=\(\frac{1006}{10085}\)
TP
15 tháng 8 2017
Ta có:
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)
\(\Rightarrow A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(\Rightarrow A=\frac{1}{1}-\frac{1}{2018}=\frac{2017}{2018}\)
\(B=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2015.2017}\)
\(\Rightarrow B=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(\Rightarrow B=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{2017}\right)=\frac{1}{2}.\frac{2016}{2017}\)
\(\Rightarrow B=\frac{1008}{2017}\)
LH
21 tháng 8 2016
Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2015.2017}\), ta có:
\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2015.2017}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)
\(=\frac{1}{2}.\frac{2016}{2017}=\frac{1008}{2017}\)
21 tháng 8 2016
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2015.2017}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2015}-\frac{1}{2017}+\frac{1}{2017}\)
\(=1-\frac{1}{2017}\)
\(=\frac{2016}{2017}\)
mk đầu tiên đấy
VN
13 tháng 5 2017
- Tính A trước:
\(A=\frac{\left(2015+1\right).2017-2}{2015+2015.2017}\)
\(A=\frac{2017.2015+2017-2}{2017.2015+2015}\)
\(A=\frac{2017.2015+2015}{2017.2015+2015}\)
\(A=1\)
- Tính B:
\(B=\frac{-2016.20172017}{2017.20162016}\)
\(B=\frac{-2016}{2017}.\frac{20172017}{20162016}\)
\(B=\frac{-10001}{10001}\)
\(B=-1\)
Vậy ta có: \(A+B=1-1=0\)
24 tháng 3 2017
A. Đặt A= biểu thức đã cho
=>\(\frac{A}{3}=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
=>\(\frac{A}{3}.2=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\)
=>\(\frac{2A}{3}-\frac{A}{3}=2-\frac{1}{2^9}\)
=>\(A=\frac{3\left(2^{10}-1\right)}{2^9}\)
B. Đặt B=biểu thức đã cho
\(\Rightarrow\frac{B}{2}=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{2015.2017}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\)
\(=\frac{1}{3}-\frac{1}{2017}=\frac{2014}{6051}\)
\(\Rightarrow B=\frac{4028}{6051}\)
31 tháng 3 2017
M = \(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{2015.2017}\)4/1.3 + 4/3.5 + 4/5.7 + ... + 4/2015.2017
M = \(2.\frac{2}{1.3}+2.\frac{2}{3.5}+2.\frac{2}{5.7}+...+2.\frac{2}{2015.2017}\) 2 . 2/1.3 + 2 . 2/3.5 + 2 . 2/5.7 + ... + 2 . 2/2015.2017
M = 2 . ( 2/1.3 + 2/3.5 + 2/5.7 + ... + 2/2015.2017 )
M = 2 . ( 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/2015 - 1/2017 )
M = 2 . ( 1 - 1/2017 )
M = 2 . 2016/2017
M = 4032/2017
31 tháng 3 2017
\(M=2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(M=2\left(1-\frac{1}{2017}\right)\)
\(M=\frac{4032}{2017}\)
\(\frac{2015.2017+1}{2015.2017}=\frac{2015.2017}{2015.2017}+\frac{1}{2015.2017}=1+\frac{1}{2015.2017}=\frac{2}{2015.2017}\)