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\(S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(S=1-\frac{1}{2018}\)
\(S=\frac{2018}{2018}-\frac{1}{2018}\)
\(S=\frac{2017}{2018}\)
\(S=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2017.2018}.\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{2017}+\frac{1}{2017}-\frac{1}{2018}\)
\(=1-\frac{1}{2018}=\frac{2017}{2018}\)
Bài làm:
Ta có: \(\frac{x}{1.2}+\frac{x}{2.3}+\frac{x}{3.4}+...+\frac{x}{2017.2018}=-1\)
\(\Leftrightarrow x\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\right)=-1\)
\(\Leftrightarrow x\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\right)=-1\)
\(\Leftrightarrow x\left(1-\frac{1}{2018}\right)=-1\)
\(\Leftrightarrow x.\frac{2017}{2018}=-1\)
\(\Rightarrow x=-\frac{2018}{2017}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2009}-\dfrac{1}{2010}\\ =1-\dfrac{1}{2010}=\dfrac{2009}{2010}\)
Đặt A , ta có :
\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{999\times1000}+1\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}+1\)
\(A=2-\frac{1}{1000}\)
\(A=\frac{2000}{1000}-\frac{1}{1000}\)
\(A=\frac{1999}{1000}\)
Đặt \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}+1=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}+1\)
\(A=1-\frac{1}{1000}+1=\frac{999}{1000}+1=\frac{1999}{1000}\)
Vậy \(A=\frac{1999}{1000}\)
Giải:
Ta có: 1/1x2+1/2x3+1/3x4+...+1/999x1000+1
= 1 - 1/2 + 1/2-1/3 + 1/3-1/4 + ...+ 1/999 - 1/1000 + 1
= 1 - 1/1000 + 1
= 2 - 1/1000
= 1999/1000
Ai tích mk mk sẽ tích lại
Ko đc Coppy
CHỉ đc viết thui nha mk cho 1 tích
vì 1/1*2=1-1/2
1/2*3=1/2-1/3
.....................
1/2014*2015=1/2014-1/2015
=1-1/2+1/2-1/3+1/3-....+1/2014-1/2015
=1-1/2015
=2014/2115
\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+....+\frac{1}{2014x2015}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\)
=\(1-\frac{1}{100}\)
=\(\frac{99}{100}\)
\(S=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2017.2018}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2017}-\dfrac{1}{2018}\)
\(=1-\dfrac{1}{2018}\)
\(=\dfrac{2017}{2018}\)