a) \(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2019\cdot2020}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(A=\frac{1}{1}-\frac{1}{2020}\)
\(A=\frac{2020}{2020}-\frac{1}{2020}=\frac{2019}{2020}\)
b) \(\frac{1}{m-1}-\frac{1}{m}=\frac{m}{m\left(m-1\right)}-\frac{m-1}{m\left(m-1\right)}=\frac{m-\left(m-1\right)}{m\left(m-1\right)}=\frac{m-m+1}{m\left(m-1\right)}=\frac{1}{m\left(m-1\right)}=\frac{1}{m^2-m}\)
Dễ dàng nhận thấy \(m^2>m^2-m\)
\(\Rightarrow\frac{1}{m^2}< \frac{1}{m^2-m}\)hay \(\frac{1}{m^2}< \frac{1}{m-1}-\frac{1}{m}\)( với m thuộc N , m > 1 ) ( đpcm )
a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(=1-\frac{1}{2020}=\frac{2019}{2020}\)
Vậy \(A=\frac{2019}{2020}\).
b) \(\frac{1}{m-1}-\frac{1}{m}=\frac{m}{m\left(m-1\right)}-\frac{m-1}{m\left(m-1\right)}=\frac{m-\left(m-1\right)}{m^2-m}=\frac{1}{m^2-m}\)
Mà \(m\inℕ;m>1\) nên \(m^2>m^2-m\)
\(\Rightarrow\frac{1}{m}< \frac{1}{m-1}-\frac{1}{m}\) (đpcm)
a, \(A=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{2019\times2020}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(=1-\frac{1}{2020}\)
\(=\frac{2019}{2020}\)
b, Ta có :
\(\frac{1}{m-1}-\frac{1}{m}=\frac{m}{m^2-m}-\frac{m-1}{m^2-m}=\frac{m-m+1}{m^2-1}=\frac{1}{m^2-1}>\frac{1}{m^2}\)
=> Đpcm
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(=1-\frac{1}{2020}=\frac{2019}{2020}\)
