Tính nhanh:
\(A=\frac{12}{1\cdot3}+\frac{12}{3\cdot5}+...+\frac{12}{2015\cdot2017}\)
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Những câu hỏi liên quan
LM
12 tháng 7 2017
a,\(\frac{2015.2016+2015-1}{2014+2015.2016}=\frac{2015.2016+2014}{2014+2015.2016}=1\)\(1\)
b,\(=1-\frac{1}{5}+\frac{1}{5}...-\frac{1}{2011}+\frac{1}{2011}-\frac{1}{2015}=1-\frac{1}{2015}=\frac{2014}{2015}\)
c,\(=\frac{12}{35}+\frac{12}{35}+\frac{12}{35}+\frac{12}{35}=\frac{12}{35}.4=\frac{48}{35}\)
25 tháng 4 2017
C=(1+1+1+...+1)+(1/1*3+1/2*4+1/3*5+...+1/2015*2017+1/2015*2017)
C=2015+(2/1*3+2/2*4+2/3*5+...+2/2015*2017+2/2015*2017):2
C=2015+(1-1/3+1/2-1/4+...+1/2015-1/2017+1/2015-1/2017):2
C=2015+(1+1/2-1/2016-1/2017+1/2015-1/2017)
cai nay thi ban tu tinh lay
nho k cho minh voi nhe
LH
29 tháng 6 2018
B = \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2015.2017}\)
B = \(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
B = \(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{2017}\right)\)
B = \(\frac{1}{2}\left(\frac{2017}{2017}-\frac{1}{2017}\right)\)
B = \(\frac{1}{2}.\frac{2016}{2017}\)
B = \(\frac{1008}{2017}\)
Vậy B = \(\frac{1008}{2017}\)
Chúc bạn học tốt . Có bài gì khó mik sẽ giúp bạn ( Chỉ toán 6 hoặc 7 trở xuống thui đó )
29 tháng 6 2018
\(B=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{2015\cdot2017}\)
\(B=\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{2015\cdot2017}\right)\)
\(B=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(B=\frac{1}{2}\left(1-\frac{1}{2017}\right)\)
\(B=\frac{1}{2}\cdot\frac{2016}{2017}\)
\(B=\frac{1008}{2017}\)
2 tháng 10 2017
A =
A = \(1-\frac{1}{2018}\)
A = \(\frac{2017}{2018}\)
Có :
2.B = \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2015.2017}\)
2.B = \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\)
2.B = \(1-\frac{1}{2017}\)
2.B = \(\frac{2016}{2017}\)
B = \(\frac{2016}{2017}:2=\frac{1008}{2017}\)
Có :
3.C = \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{2017.2020}\)
3.C = \(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{2017}-\frac{1}{2020}\)
3.C = \(\frac{1}{1}-\frac{1}{2020}=\frac{2019}{2020}\)
C = \(\frac{2019}{2020}:3=\frac{673}{2020}\)
BH
25 tháng 3 2019
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+.....+\frac{1}{2016\cdot2017}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.........+\frac{1}{2016}-\frac{1}{2017}\)
\(=1-\frac{1}{2017}=\frac{2016}{2017}\)
25 tháng 3 2019
= 1/1-1/2+1/2-1/3+1/3-............-1/2017
=1-1/2017
=2016/2017
8 tháng 7 2018
\(\frac{4}{3.5}-\frac{6}{5.7}+\frac{8}{7.9}+\frac{10}{9.11}+...+\frac{2016}{2015.2017}\)
\(=2.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(=2.\left(\frac{1}{3}-\frac{1}{2017}\right)\)
\(=2.\frac{2014}{6051}\)
\(=\frac{4028}{6051}\)
\(\Rightarrow BT>\frac{1}{6}\)
NM
1
TM
26 tháng 9 2017
Đặt \(A=1.2.3+2.3.4+3.4.5+...+2015.2016.2017\)
=>\(4A=1.2.3.4+2.3.4.4+3.4.5.4+...+2015.2016.2017.4\)
=>\(4A=1.2.3.\left(4-0\right)+2.3.4.\left(5-1\right)+3.4.5.\left(6-2\right)\)
\(+...+2015.2016.2017.\left(2018-2014\right)\)
=>\(4A=1.2.3.4-0.1.2.3+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5\)
\(+...+2015.2016.2017.2018-2014.2015.2016.2017\)
=>\(4A=2015.2016.2017.2018\Rightarrow A=\frac{2015.2016.2017.2018}{4}\)
A = 6.(1/1.3+1/3.5+...+1/2015.2017)
= 6.(1/1-1/3+1/3-1/5+...+1/2015-1/2017)
= 6.(1/1-1/2017)
= 6.2016/2017
=12096/2017