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Ta có:
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)
\(\Rightarrow A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(\Rightarrow A=\frac{1}{1}-\frac{1}{2018}=\frac{2017}{2018}\)
\(B=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2015.2017}\)
\(\Rightarrow B=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(\Rightarrow B=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{2017}\right)=\frac{1}{2}.\frac{2016}{2017}\)
\(\Rightarrow B=\frac{1008}{2017}\)
a)1/1x2+1/2x3+....+1/2003x2004
=1-1/2+1/2-1/3+...+1/2003+1/2004
=1-1/2004
=2004/2004-1/2004
=2003/2004
b)1/1x3+1/3x5+...+1/2003x2005
=1-1/3+1/3-1/5+....+1/2003+1/2005
=1-1/2005
=2005/2005-1/2005
=2004/2005
a) \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{2003\cdot2004}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2003}-\frac{1}{2004}\)
\(=1-\frac{1}{2004}=\frac{2003}{2004}\)
b) Đặt A=\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{2003\cdot2005}\)
\(2A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{1}{5\cdot7}+....+\frac{2}{2003\cdot2005}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2003}-\frac{1}{2005}\)
\(2A=1-\frac{1}{2005}\)
\(2A=\frac{2004}{2005}\)
\(A=\frac{2004}{2005}:2=\frac{2004}{2005}\cdot\frac{1}{2}=\frac{1002}{2005}\)
a)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2003.2004}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2003}-\frac{1}{2004}\)
\(=\frac{1}{1}-\frac{1}{2004}\)
\(\Rightarrow=\frac{2003}{2004}\)
b)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2003+2005}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2003}-\frac{1}{2005}\)
\(=\frac{1}{1}-\frac{1}{2005}\)
\(\Rightarrow=\frac{2004}{2005}\)
Ta có:
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2003.2004}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2003}-\frac{1}{2004}\)
\(=1-\frac{1}{2004}=\frac{2003}{2004}\)
b,
\(\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2003.2005}\right).\frac{1}{2}\)
\(\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2003}-\frac{1}{2005}\right).\frac{1}{2}\)
\(=\left(1-\frac{1}{2005}\right).\frac{1}{2}=\frac{2004}{2005}.\frac{1}{2}=\frac{1002}{2005}\)
Nhớ nha bạn
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(A=1-\frac{1}{2017}\)
\(A=\frac{2016}{2017}\)
Sao k có ai giúp mk hết vậy >:((, thôi để mk tự giúp mk vậy :>. E mới nghĩ ra cách này có gì sai anh giúp đỡ.
Cách 1 - Ta có :
\(A=\frac{1}{1.2}+\frac{1}{1.3}+\frac{1}{1.4}+...+\frac{1}{3.2}+\frac{1}{3.3}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{6}+\frac{1}{9}\)
\(\Rightarrow A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{6}+\frac{1}{9}\)
\(\Rightarrow A=\frac{5}{6}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{6}+\frac{1}{9}\)
Mà \(\frac{5}{6}>\frac{2}{3}\Rightarrow\frac{5}{6}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{6}+\frac{1}{9}>\frac{2}{3}\)
\(\Leftrightarrowđpcm\)
A = 1 - 1/2 + 1/2 - 1/3 + .... + 1/43 - 1/44 + 1/44 - 1/45 = 1 - 1/45 = 44/45
Theo mình nghĩ thì đề bải phải là 1/2.3 chứ không phải là 1/1.3
k nha