\(=\dfrac{1}{100}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{100}-\dfrac{99}{100}=-\dfrac{49}{50}\)
\(=\dfrac{1}{100}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{100}-\dfrac{99}{100}=-\dfrac{49}{50}\)
\(\dfrac{-1}{99}\dfrac{1}{99.98}-\dfrac{1}{98.97}-\dfrac{1}{97.96}-...\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
B1
(-1/2)-(-3/5)+(-1/9)+1/131-(-2/7)+4/35-7/18
B2
A=1/3-3/5+5/7-7/9+9/11-11/13+13/15+11/13-9/11+7/9-5/7+3/5-1/3
B=1/99-1/99.98-1/98.97-1/97.96-...-1/3.2-1/2.1
a) \(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{13}+\dfrac{13}{15}+\dfrac{11}{13}-\dfrac{9}{11}+\dfrac{7}{9}\)\(-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{1}{3}\)
b)\(\dfrac{1}{99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-\dfrac{1}{97.96}-.....-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
c) \(\dfrac{1}{1.3}+\dfrac{1}{3.5}+.....+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}+......+\dfrac{1}{255.257}\)
Tính :
a) A= \(\dfrac{1}{2010.2009}-\dfrac{1}{2009.2008}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
b) B= 63.(-124)-124.(-37)+(\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{11}\)).(-66)
Giúp mình với ạ! Mình cần gấp lắm
Cho biểu thức:
P=\(\frac{1}{2000.1999}-\frac{1}{1999.1998}-\frac{1}{1998.1997}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
Kết quả của phép tính P+\(\frac{1997}{1998}\)
\(\dfrac{1}{50}-\dfrac{1}{50.49}-\dfrac{1}{49.48}-...-\dfrac{1}{2.1} \)
Tính các số hữu tỉ sau:
A=\(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right).....\left(1-\dfrac{1}{n+1}\right),n\)là số tự nhiên
B=\(\dfrac{1}{2000.1999}-\dfrac{1}{1999.1998}-\dfrac{1}{1998.1997}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
Nhờ các các bn giúp mik!khó quá mik lm hk ra ai giỏi toán giúp mik với!
\(\dfrac{1^2.a^3}{a^2.1^2}\)
B= \(\left[\left(0,1\right)^2\right]^0\) +\(\left[\left(\frac{1}{7}\right)^{-1}\right]^2\) .\(\frac{1}{49}.\left[\left(2^2\right)^3.2^5\right]\)