\(\frac{1}{1x2}+\frac{1}{2x3}+...+\frac{1}{2018+2019}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2018}-\frac{1}{2019}\)
\(=1-\frac{1}{2019}\)
\(=\frac{2018}{2019}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2018.2019}\) ( đúng ko bn ?? )
= \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2018}-\frac{1}{2019}\)
= \(\frac{1}{1}-\frac{1}{2019}=\frac{2018}{2019}\)
Học tốt
Đề bài sai à :v
Đặt \(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{2018\cdot2019}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2018}-\frac{1}{2019}\)
\(A=\frac{1}{1}-\frac{1}{2019}\)
\(A=\frac{2018}{2019}\)
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