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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 = \(\frac{2019}{2020}:3=\frac{673}{2020}\)
\(\frac{1}{2}.\left(1+\frac{1}{1.3}\right).\left(\frac{1}{2.4}+1\right).....\left(1+\frac{1}{2015.2017}\right)\)
\(=\frac{1}{2}.\left(\frac{4}{1.3}.\frac{9}{2.4}......\frac{4064256}{2015.2017}\right)\)
\(=\frac{1}{2}.\left(\frac{2.2}{1.3}.\frac{3.3}{2.4}.....\frac{2016.2016}{2015.2017}\right)\)
\(=\frac{1}{2}.\left(\frac{2.3....2016}{1.2....2015}.\frac{2.3.....2016}{3.4....2017}\right)\)
\(=\frac{1}{2}.2016.\frac{2}{2017}\)
\(=1008.\frac{2}{2017}=\frac{2016}{2017}\)
a,(1/3/7-2/1/4) . 3/1/3
= -23/28 .3/1/3
= -115/42
b,(2/1/3+3/1/2):(-4/1/6+3/1/7)+7/1/2
= 35/6 : -43/42 +7/1/2
= -245/43 +7/1/2
= 155/86
\(\left(1\frac{3}{7}-2\frac{1}{4}\right).3\frac{1}{3}\)
\(=\left(\frac{10}{7}-\frac{9}{4}\right).\frac{10}{3}\)
\(=-\frac{23}{28}.\frac{10}{3}\)
\(=\frac{-115}{42}\)
Đặt \(A=1.2.3+2.3.4+3.4.5+...+2015.2016.2017\)
=>\(4A=1.2.3.4+2.3.4.4+3.4.5.4+...+2015.2016.2017.4\)
=>\(4A=1.2.3.\left(4-0\right)+2.3.4.\left(5-1\right)+3.4.5.\left(6-2\right)\)
\(+...+2015.2016.2017.\left(2018-2014\right)\)
=>\(4A=1.2.3.4-0.1.2.3+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5\)
\(+...+2015.2016.2017.2018-2014.2015.2016.2017\)
=>\(4A=2015.2016.2017.2018\Rightarrow A=\frac{2015.2016.2017.2018}{4}\)
\(\frac{4}{3.5}-\frac{6}{5.7}+\frac{8}{7.9}+\frac{10}{9.11}+...+\frac{2016}{2015.2017}\)
\(=2.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(=2.\left(\frac{1}{3}-\frac{1}{2017}\right)\)
\(=2.\frac{2014}{6051}\)
\(=\frac{4028}{6051}\)
\(\Rightarrow BT>\frac{1}{6}\)
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 )
\(B=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{2015\cdot2017}\)
\(B=\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{2015\cdot2017}\right)\)
\(B=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(B=\frac{1}{2}\left(1-\frac{1}{2017}\right)\)
\(B=\frac{1}{2}\cdot\frac{2016}{2017}\)
\(B=\frac{1008}{2017}\)