A=\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+...+\(\dfrac{1}{99.100}\)
A=\(\dfrac{1}{1}\)-\(\dfrac{1}{2}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+...+\(\dfrac{1}{99}\)-\(\dfrac{1}{100}\)
A=\(\dfrac{1}{1}\)-\(\dfrac{1}{100}\)=\(\dfrac{100}{100}\)-\(\dfrac{1}{100}\)=\(\dfrac{99}{100}\)
Vậy: A=\(\dfrac{99}{100}\)
Chứng minh rằng:
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\)
Tính tổng
=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
Tính A:
A=\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)
HELP ME!
tính nhanh
A=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}\)
B=\(\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{120}\)
các bn giúp mk nha
Tính: \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2017.2018}+\dfrac{1}{2018.2019}\)
Chứng minh rằng:
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
1, Tính
a, B=\(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + . . . + \(\dfrac{1}{2007.2008}\)
b, Q= \(\dfrac{7}{1.3}\) + \(\dfrac{7}{3.5}\) + . . . + \(\dfrac{7}{2009.2011}\)
c, S= \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\)+ . . . + \(\dfrac{1}{3^{5000}}\)
1,Tính
a, A = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + . . . + \(\dfrac{1}{2017.2018}\)
b,B = \(\dfrac{5}{2.4}\) + \(\dfrac{5}{4.6}\) + \(\dfrac{5}{6.8}\) + . . . + \(\dfrac{5}{2016.2018}\)
c. C = \(\dfrac{1}{18}\) + \(\dfrac{1}{54}\) + \(\dfrac{1}{108}\) + . . . + \(\dfrac{1}{990}\)
Tính :
\(\left(1-\dfrac{2}{2.3}\right).\left(1-\dfrac{2}{3.4}\right).\left(1-\dfrac{2}{4.5}\right).....\left(1-\dfrac{2}{99.100}\right)\)
Tìm x biết :
\(\dfrac{1}{x}-\dfrac{y}{6}=\dfrac{1}{3}\)
m.n giúp mk nha ! 