Tính: \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2017.2018}+\dfrac{1}{2018.2019}\)
\(\dfrac{x}{1.2}+\dfrac{x}{2.3}+\dfrac{x}{3.4}+...+\dfrac{x}{2017.2018}=-1\)
`x/(1.2)+x/(2.3)+x/(3.4)+.....+x/(2017.2018)=1`
`-> x/1 - x/2 +x/2-x/3+x/3-x/4+........+x/2017-x/2018=1`
`-> x-x/2018=1`
`-> 2017/2018 .x=1`
`-> x=2018/2017`
Tính tổng
$\dfrac{1}{1.2}$ + $\dfrac{1}{2.3}$ + $\dfrac{1}{3.4}$ + .... + $\dfrac{1}{99.100}$
= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100
= 1/1 - 1/100
= 99/100
Học từ lớp 4 rồi :V
tính nhanh
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}=?????\)
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}\)
=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}\)
=\(1-\dfrac{1}{5}\)
=\(\dfrac{4}{5}\)
1/1 - 1/2+ 1/2 - 1/3 + 1/3 - 1/4+ 1/4 - 1/5
= 1/1 - 1/5
= 4/5
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}=1-\dfrac{1}{5}=\dfrac{4}{5}\)
Chứng minh rằng : \(\dfrac{A}{B}\in Z\) với :
\(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{2015.2016}+\dfrac{1}{2017.2018};\)
\(B=\dfrac{1}{1010.2018}+\dfrac{1}{1011.2017}+...+\dfrac{1}{2018.1010}\).
Ta biến đổi \(A=\dfrac{2-1}{1.2}+\dfrac{4-3}{3.4}+...+\dfrac{2016-2015}{2016.2015}+\dfrac{2018-2017}{2017.2018}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}-\dfrac{1}{2016}+\dfrac{1}{2017}-\dfrac{1}{2018}\)
\(A=\left(1+\dfrac{1}{3}+...+\dfrac{1}{2017}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2018}\right)\)
\(A=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2017}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2018}\right)\)
\(A=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2017}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{1009}\right)\)
\(A=\dfrac{1}{1010}+\dfrac{1}{1011}+...+\dfrac{1}{2017}+\dfrac{1}{2018}\)
Lại có \(B=\dfrac{1}{1010.2018}+\dfrac{1}{1011.2017}+...+\dfrac{1}{2018.1010}\)
\(B=\dfrac{1}{3028}.\left(\dfrac{3028}{1010.2018}+\dfrac{3028}{1011.2017}+...+\dfrac{3028}{2018.1010}\right)\)
\(B=\dfrac{1}{3028}\left(\dfrac{1}{1010}+\dfrac{1}{2018}+\dfrac{1}{1011}+\dfrac{1}{2017}+...+\dfrac{1}{2018}+\dfrac{1}{1010}\right)\)
\(B=\dfrac{1}{3028}.2\left(\dfrac{1}{1010}+\dfrac{1}{1011}+...+\dfrac{1}{2018}\right)\)
\(B=\dfrac{1}{3028}.2A\) \(\Rightarrow\dfrac{A}{B}=1514\inℤ\). Ta có đpcm
tính nhanh:
A = \(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+........+\(\dfrac{1}{149.150}\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{149.150}\)
\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{149}-\dfrac{1}{150}\)
\(A=\dfrac{1}{1}-\dfrac{1}{150}\)
\(A=\dfrac{150}{150}-\dfrac{1}{150}\)
\(A=\dfrac{149}{150}\)
3. tính:
A=\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+...+\(\dfrac{1}{49.50}\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)
\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(A=\dfrac{1}{1}-\dfrac{1}{50}\)
\(A=\dfrac{49}{50}\)
A = 1/1.2 +1/2.3 +1/3.4 +...+1/49.50
A = 1 +1/2 -1/2+1/3-1/3+1/4-...-1/49 +1/50
A = 1 - 1/50
A=49/50
Cho \(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+....+\dfrac{1}{2017.2018}\) và \(B=\dfrac{1}{1010.2018}+\dfrac{1}{1011.2017}+...+\dfrac{1}{2018.1010}\). C/m A/B là 1 số nguyên
Tính tổng sau: \(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{999.1000}\)
Áp dụng công thức \(\dfrac{1}{k\left(k+1\right)}=\dfrac{1}{k}-\dfrac{1}{k+1}\), ta có:
\(A=\left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{999}-\dfrac{1}{1000}\right)=1-\dfrac{1}{1000}=\dfrac{999}{1000}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{999}-\dfrac{1}{1000}\)
\(A=1-\dfrac{1}{1000}\)
\(A=\dfrac{999}{1000}\)
tính tổng
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.....+\dfrac{1}{n.\left(n+1\right)}\)
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{n\left(n+1\right)}\)
= \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\)
= 1 - \(\dfrac{1}{n+1}\) = \(\dfrac{n}{n+1}\)